SK Tuitions • Class 8 Mathematics • Ganita Prakash

Number Play — 80 Question Bank

Comprehensive practice on parity, factors and multiples, remainders, algebraic number forms, divisibility tests, digital roots, consecutive-number reasoning and cryptarithms. Every question includes a detailed click-to-view solution.

80Total Questions

30One Markers

302M + 3M + 5M

20A–R + Cases

Section A — 30 One-Mark Questions

Q1 1 Mark Parity

Which of the following is always even?
(A) odd + even   (B) odd + odd   (C) odd − even   (D) odd × odd

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Answer: (B)
Let two odd integers be 2m+1 and 2n+1. Their sum is 2(m+n+1), which has factor 2 and is therefore even.

Q2 1 Mark Ganita Prakash concept

For any four consecutive integers, every expression obtained by inserting only ‘+’ or ‘−’ signs between them has the same ______.

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The blank is parity. Changing one sign changes the value by twice one of the numbers, which is an even change; therefore parity is unchanged.

Q3 1 Mark Chapter-based

Write an algebraic form for an even integer.

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Any even integer is a multiple of 2, so it can be written as 2n, where n is an integer.

Q4 1 Mark Chapter-based

Write an algebraic form for an odd integer.

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An odd integer is one more than an even integer, so it can be written as 2n + 1.

Q5 1 Mark Ganita Prakash concept

Which expression is always even for integer values of m and q?
(A) 3m+q   (B) 4m+2q   (C) m²+2   (D) 5m+1

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Answer: (B)
4m+2q=2(2m+q), so 2 is a factor of the entire expression.

Q6 1 Mark Ganita Prakash concept

An even number that is not a multiple of 4 leaves what remainder when divided by 4?

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It leaves remainder 2. Every even integer is congruent to either 0 or 2 modulo 4.

Q7 1 Mark Ganita Prakash concept

If two numbers are both multiples of 8, is their sum necessarily a multiple of 8?

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Yes. If the numbers are 8a and 8b, their sum is 8(a+b), which is a multiple of 8.

Q8 1 Mark Divisibility property

True or False: If a divides both M and N, then a also divides M−N.

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True. Write M=ap and N=aq. Then M−N=a(p−q), which is divisible by a.

Q9 1 Mark Ganita Prakash concept

If a number is divisible by 12, which of the following must divide it: 2, 3, 4, 6?

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All four. They are all factors of 12, and every multiple of 12 is divisible by every factor of 12.

Q10 1 Mark Chapter-based

The sum of an odd number and an even number is always ______.

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Odd. (2m+1)+2n=2(m+n)+1.

Q11 1 Mark Ganita Prakash concept

Write the general form of numbers that leave remainder 3 when divided by 5.

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The general form is 5k + 3, where k is an integer (or non-negative integer for positive numbers).

Q12 1 Mark Ganita Prakash concept

Which expression also generates positive integers leaving remainder 3 when divided by 5, for k≥1?

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5k−2, because 5k−2 = 5(k−1)+3.

Q13 1 Mark Ganita Prakash problem

Find the remainder when 4779+661 is divided by 7, given that 4779 leaves remainder 5 and 661 leaves remainder 3.

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Add the remainders: 5+3=8. Since 8=7+1, the remainder is 1.

Q14 1 Mark Ganita Prakash problem

Find the remainder when 4779−661 is divided by 7, using the same remainder information.

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Subtract the remainders: 5−3=2. Hence the remainder is 2.

Q15 1 Mark Ganita Prakash problem

A number leaves remainder 2 when divided by 3, remainder 3 when divided by 4, and remainder 4 when divided by 5. What is the smallest such positive number?

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Each remainder is one less than the divisor. So N+1 must be a common multiple of 3,4,5. The least is 60; hence N=59.

Q16 1 Mark Ganita Prakash concept

What is the divisibility test for 9?

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A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

Q17 1 Mark Ganita Prakash problem

Is 405 divisible by 9?

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Digit sum = 4+0+5=9, which is divisible by 9. Therefore yes.

Q18 1 Mark Ganita Prakash example

Find the digital root of 489710.

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4+8+9+7+1+0=29; 2+9=11; 1+1=2.

Q19 1 Mark Digital roots

What is the digital root of a positive multiple of 9?

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It is 9.

Q20 1 Mark Ganita Prakash concept

What is the divisibility test for 3?

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A number is divisible by 3 if the sum of its digits is divisible by 3.

Q21 1 Mark Ganita Prakash concept

For divisibility by 11, what should be true of the alternating sum of the digits?

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The alternating sum must be 0 or a multiple of 11.

Q22 1 Mark Ganita Prakash problem

Is 90904 divisible by 11?

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Alternating sum from the units side: 4−0+9−0+9=22, a multiple of 11. Therefore yes.

Q23 1 Mark Composite divisibility

To test divisibility by 6, which two divisibility tests must both be satisfied?

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The number must be divisible by 2 and 3.

Q24 1 Mark Ganita Prakash concept

To test divisibility by 24 using coprime factors, checking divisibility by which pair works: (4,6) or (3,8)?

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(3,8). Since gcd(3,8)=1 and 3×8=24, divisibility by both ensures divisibility by 24.

Q25 1 Mark Ganita Prakash problem

If 31z5 is divisible by 9, what are the possible values of z?

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Digit sum = 3+1+z+5=9+z. This must be a multiple of 9. Since z is a digit, z=0 or 9.

Q26 1 Mark Ganita Prakash problem

If a multiple of 11 is doubled, is the result always a multiple of 11?

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Yes. If n=11k, then 2n=22k=11(2k).

Q27 1 Mark Ganita Prakash problem

The product of two consecutive integers is always divisible by which smallest integer greater than 1?

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2, because one of two consecutive integers must be even.

Q28 1 Mark Ganita Prakash problem

The product of three consecutive integers is always divisible by 6. Why?

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Among any three consecutive integers, at least one is divisible by 3 and at least one is even. Hence the product contains factors 3 and 2, so it is divisible by 6.

Q29 1 Mark Ganita Prakash problem

Which set relation is correct?
(A) multiples of 4 ⊂ multiples of 8 ⊂ multiples of 32
(B) multiples of 32 ⊂ multiples of 8 ⊂ multiples of 4

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Answer: (B). Every multiple of 32 is a multiple of 8, and every multiple of 8 is a multiple of 4.

Q30 1 Mark Ganita Prakash problem

Solve the cryptarithm value: if UT×3=PUT, the PDF solution gives U=5, T=0. What is P?

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50×3=150, so P=1.

Section B — 10 Two-Mark Questions

Q31 2 Marks Ganita Prakash concept

Show algebraically that switching +b to −b in a+b−c−d does not change the parity of the expression.

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Original expression E₁=a+b−c−d.
New expression E₂=a−b−c−d.
E₁−E₂=(a+b−c−d)−(a−b−c−d)=2b.
Since 2b is even, E₁ and E₂ differ by an even number and therefore have the same parity.

Q32 2 Marks Ganita Prakash concept

When is the sum of two even numbers a multiple of 4? Give the three remainder cases modulo 4.

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Every even number is either 4p or 4p+2.
Case 1: 4p+4q=4(p+q), divisible by 4.
Case 2: (4p+2)+(4q+2)=4(p+q+1), divisible by 4.
Case 3: 4p+(4q+2)=4(p+q)+2, not divisible by 4.
Thus the sum is a multiple of 4 when both even numbers are of the same type modulo 4.

Q33 2 Marks Ganita Prakash concept

Classify: ‘If a number is divisible by 7, then it is divisible by any multiple of 7.’ Give one example and one counterexample.

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Sometimes true.
Example: 42 is divisible by 7 and by 14.
Counterexample: 42 is divisible by 7 but not by 28.
A number 7k is divisible by 7m only when m divides k.

Q34 2 Marks Ganita Prakash problem

Write all positive integers that leave remainder 2 when divided by both 3 and 4 in the form an+b.

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If x leaves remainder 2 in both cases, then x−2 is divisible by both 3 and 4. Therefore x−2 is a multiple of LCM(3,4)=12.
So x=12n+2. Examples: 14, 26, 38, 50, …

Q35 2 Marks Ganita Prakash example

Without dividing, find the remainder when 7309 is divided by 9.

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For division by 9, a number and its digit sum have the same remainder.
7+3+0+9=19; 19 leaves remainder 1 on division by 9.
Hence the remainder is 1.

Q36 2 Marks Ganita Prakash problem

Use the divisibility test for 11 to decide whether 857076 is divisible by 11.

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Take alternating sum from units side: 6−7+0−7+5−8 = −11.
Since −11 is a multiple of 11, 857076 is divisible by 11.

Q37 2 Marks Ganita Prakash concept

Explain why checking divisibility by 4 and 6 is not sufficient to prove divisibility by 24. Give a counterexample.

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The number 12 is divisible by both 4 and 6, but 12 is not divisible by 24. The two tests overlap in a factor 2; together they only guarantee the LCM of 4 and 6, which is 12, not 24.

Q38 2 Marks Ganita Prakash problem

Find the digital root of 9a+36b+13 for integer values of a and b.

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9a+36b+13 = 9(a+4b+1)+4.
The first part is a multiple of 9, so the number leaves remainder 4 modulo 9. Therefore its digital root is 4.

Q39 2 Marks Ganita Prakash problem

Three numbers each leave remainder 2 when divided by 6. Show that their sum is a multiple of 6.

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Let the numbers be 6a+2, 6b+2 and 6c+2.
Their sum = 6(a+b+c)+6 = 6(a+b+c+1), which is a multiple of 6.

Q40 2 Marks Ganita Prakash problem

If the greatest of five consecutive integers is p, write the other four and find their sum in terms of p.

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The five integers are p−4, p−3, p−2, p−1, p.
The other four are p−1, p−2, p−3, p−4.
Their sum = 4p−(1+2+3+4)=4p−10.

Section C — 10 Three-Mark Questions

Q41 3 Marks Ganita Prakash problem

The sum of four consecutive integers is 34. Find the integers algebraically.

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Let the integers be n, n+1, n+2, n+3.
n+(n+1)+(n+2)+(n+3)=34
4n+6=34 ⇒ 4n=28 ⇒ n=7.
Therefore the integers are 7, 8, 9, 10.

Q42 3 Marks Ganita Prakash problem

Classify and justify: ‘The sum of a multiple of 6 and a multiple of 9 is always a multiple of 3.’

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Always true.
Let the numbers be 6x and 9y.
Their sum = 6x+9y = 3(2x+3y).
Since the sum has factor 3 for all integers x,y, it is always divisible by 3.

Q43 3 Marks Ganita Prakash problem

Find the smallest positive integer that leaves remainder 1 when divided by 3 and 5, is odd, is divisible by 7, and is less than 100.

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Remainder 1 on division by 3 and 5 means N−1 is a multiple of LCM(3,5)=15. Oddness plus the same condition suggests N=30k+1, because 30k is even and N is odd.
Candidates below 100: 31, 61, 91.
Only 91 is divisible by 7. Hence N=91.

Q44 3 Marks Ganita Prakash problem

Find the smallest multiple of 9 containing no odd digit, and explain why a two-digit answer is impossible.

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Allowed non-zero digits are 2,4,6,8 (and 0 may occur). A number divisible by 9 must have digit sum divisible by 9.
The sum of two even digits is even, so it cannot equal 9; the next possible positive multiple of 9 is 18.
The smallest multiset of allowed digits summing to 18 that forms the smallest number is 2,8,8.
Thus the smallest number is 288.

Q45 3 Marks Ganita Prakash problem

Find the closest multiple of 9 to 6000 and show the reasoning without long division.

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Digit sum of 6000 is 6, so 6000 leaves remainder 6 on division by 9.
The previous multiple is 6000−6=5994, distance 6.
The next multiple is 6000+(9−6)=6003, distance 3.
Hence the closest multiple is 6003.

Q46 3 Marks Ganita Prakash problem

How many multiples of 9 lie strictly between 4300 and 4400?

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4300 has digit sum 7, so the next multiple of 9 is 4302. The last multiple below 4400 is 4392.
Sequence: 4302, 4311, …, 4392.
Number of terms = (4392−4302)/9 + 1 = 90/9 + 1 = 11.

Q47 3 Marks Ganita Prakash concept

A number is divisible by both 9 and 4. Prove that it must be divisible by 36.

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Because gcd(9,4)=1, their LCM is 9×4=36.
If N is divisible by both 9 and 4, its prime factorisation contains 3² and 2². Therefore it contains 2²·3²=36 as a factor.
Hence N is divisible by 36.

Q48 3 Marks Ganita Prakash concept

A number is divisible by both 6 and 4. Must it be divisible by 24? Classify the statement and justify.

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Sometimes true.
Being divisible by 6 and 4 guarantees divisibility by LCM(6,4)=12, not necessarily 24.
Counterexample: 12 is divisible by both 6 and 4 but not 24.
Example: 24 is divisible by 6, 4 and 24.

Q49 3 Marks Ganita Prakash problem

Explain why reversing the digits of a multiple of 9 produces another multiple of 9.

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The divisibility test for 9 depends only on the sum of the digits. Reversing the digits does not change that sum.
If the original digit sum is 9k, the reversed number has the same digit sum 9k, so it is also divisible by 9.
The same reasoning works for any permutation of the digits.

Q50 3 Marks Ganita Prakash problem

If the middle number of five consecutive even integers is 5p, express the full sequence and find its sum.

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Consecutive even integers differ by 2. Therefore the sequence is:
5p−4, 5p−2, 5p, 5p+2, 5p+4.
Adding, the ±4 and ±2 terms cancel, giving total = 25p.

Section D — 10 Five-Mark Questions

Q51 5 Marks Ganita Prakash concept

Prove that for any four integers a,b,c,d, all eight expressions a±b±c±d have the same parity. Give two different reasoning methods.

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Method 1: Sign-switch argument.
Start with one expression, say a+b−c−d. Changing +b to −b changes the value by 2b; changing −c to +c changes it by 2c, and similarly for any sign. Every sign change changes the value by an even number, so parity never changes. Since any of the eight expressions can be reached by sign changes, all have the same parity.

Method 2: Parity rules.
For any two integers x,y, x+y and x−y have the same parity. Applying this repeatedly shows that replacing any + by − does not alter parity. Hence every expression a±b±c±d has the same parity.

Q52 5 Marks Ganita Prakash concept

Investigate the statement: ‘If a number is divisible by k and by m, then it is divisible by km.’ State the correct general rule and illustrate it with (9,4) and (6,4).

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The statement is not always true because k and m may share factors.
The correct rule is: if N is divisible by both k and m, then N is divisible by LCM(k,m).

For 9 and 4: gcd(9,4)=1, so LCM=36=9×4. Thus divisibility by both 9 and 4 does imply divisibility by 36.
For 6 and 4: LCM(6,4)=12, not 24. For example, 12 is divisible by both 6 and 4 but not 24.
Therefore multiplication of the two divisors is valid only when the divisors are coprime.

Q53 5 Marks Ganita Prakash concept

Derive the divisibility rule for 9 using place value. Explain why the digit sum has the same remainder as the number.

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Write a number N with digits …dcba as …+1000d+100c+10b+a.
Since 10=9+1, 100=99+1, 1000=999+1, and in general 10^r is one more than a multiple of 9, each place value contributes its digit plus a multiple of 9.
Thus N = 9K + (a+b+c+d+…).
So N and its digit sum have the same remainder modulo 9. Therefore N is divisible by 9 exactly when its digit sum is divisible by 9.
Example: 7309 has digit sum 19; 19 leaves remainder 1 modulo 9, so 7309 also leaves remainder 1.

Q54 5 Marks Ganita Prakash concept

Derive and apply the divisibility rule for 11 to the number 320185.

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Powers of 10 alternate modulo 11: 1≡+1, 10≡−1, 100≡+1, 1000≡−1, and so on.
Therefore a number is congruent modulo 11 to an alternating sum of its digits.
For 320185, taking + on the units position gives: 5−8+1−0+2−3 = −3.
So the number is 3 short of a multiple of 11, equivalently it leaves remainder 8 when divided by 11.
A number is divisible by 11 when the alternating sum is 0 or a multiple of 11.

Q55 5 Marks Ganita Prakash problem

If 48a23b is divisible by 18, find all possible digit pairs (a,b) and explain all conditions.

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Divisibility by 18 requires divisibility by 2 and 9.
So b must be even: b∈{0,2,4,6,8}.
Digit sum = 4+8+a+2+3+b = 17+a+b. This must be a multiple of 9.
Since 17+a+b ranges from 17 to 34, possible multiples are 18 or 27.
For b=0, a=1; b=2, a=8; b=4, a=6; b=6, a=4; b=8, a=2.
Hence the pairs are (1,0), (8,2), (6,4), (4,6), (2,8).

Q56 5 Marks Ganita Prakash problem

If 3p7q8 is divisible by 44, determine all possible digit pairs (p,q).

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Since 44=4×11, the number must be divisible by 4 and 11.
Divisibility by 4: last two digits q8 must be divisible by 4, so q∈{0,2,4,6,8}.
Divisibility by 11: alternating sums are 8+7+3=18 and p+q, so 18−(p+q) must be 0 or ±11.
p+q=18 is impossible with the allowed q and digit p under the given constraints; 18−(p+q)=11 gives p+q=7.
Thus: q=0,p=7; q=2,p=5; q=4,p=3; q=6,p=1. q=8 would require p=−1.
Pairs: (7,0), (5,2), (3,4), (1,6).

Q57 5 Marks Ganita Prakash problem

Find and generalise all triples of consecutive integers in which the first is divisible by 2, the second by 3, and the third by 4.

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Let the triple be n,n+1,n+2.
We require n≡0 (mod 2), n+1≡0 (mod 3), n+2≡0 (mod 4).
One solution is n=2, giving 2,3,4.
The combined pattern repeats modulo LCM(2,3,4)=12. Therefore n=2+12k.
Hence all triples are (12k+2, 12k+3, 12k+4), for integers k≥0. Examples: 2,3,4; 14,15,16; 26,27,28; …

Q58 5 Marks Ganita Prakash problems

Explain the digital root patterns for multiples of 3, 4 and 6. Relate digital roots to remainders modulo 9.

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For any positive integer, the digital root is the representative from 1 to 9 having the same remainder modulo 9 (with remainder 0 represented by 9).
Multiples of 3 therefore have digital roots repeating 3,6,9.
Multiples of 4 advance by 4 modulo 9, giving 4,8,3,7,2,6,1,5,9, then repeating.
Multiples of 6 advance by 6 modulo 9, giving 6,3,9, then repeating.
Thus digital-root patterns are modular patterns, not simply parity patterns.

Q59 5 Marks Ganita Prakash problem

Prove the divisibility results for products of 2, 3, 4 and 5 consecutive integers.

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For two consecutive integers, one is even, so their product is divisible by 2.
Among three consecutive integers, one is divisible by 3 and at least one by 2, so the product is divisible by 6=3!.
Among four consecutive integers, there are two even numbers, one of which is divisible by 4, and one number divisible by 3. Hence the product contains factors 4×2×3=24 and is divisible by 24=4!.
Among five consecutive integers, the product contains all factors required for 5!=120, so it is divisible by 120.
This is a special case of the fact that the product of n consecutive integers is divisible by n!.

Q60 5 Marks Ganita Prakash problem

A 6-digit number is divisible by 15, and when its digits are reversed the resulting 6-digit number is divisible by 6. Derive practical conditions and give one valid example.

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Original divisibility by 15 means divisibility by 3 and 5. Since reversing must still give a 6-digit number, the last digit of the original cannot be 0; therefore it must be 5.
The reversed number must be divisible by 6, so it must be divisible by 2 and 3. Its last digit is the original first digit, so the original first digit must be even: 2,4,6 or 8.
Digit sum is unchanged by reversal, so choosing a digit sum divisible by 3 satisfies the divisibility-by-3 condition for both numbers.
Example: 200025. It ends in 5 and digit sum is 9, so it is divisible by 15. Reversed: 520002, which is even and has digit sum 9, so it is divisible by 6.

Section E — 10 Assertion–Reason Questions

Q61 1 Mark Chapter-based

For Assertion–Reason questions use: (A) Both A and R are true and R correctly explains A; (B) Both are true but R is not the correct explanation; (C) A is true but R is false; (D) A is false but R is true.

Assertion (A): The sum of two odd integers is even.
Reason (R): (2m+1)+(2n+1)=2(m+n+1).

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Answer: (A). The reason directly proves the assertion.

Q62 1 Mark Chapter-based

For Assertion–Reason questions use: (A) Both A and R are true and R correctly explains A; (B) Both are true but R is not the correct explanation; (C) A is true but R is false; (D) A is false but R is true.

Assertion (A): If two numbers are multiples of 8, their difference is a multiple of 8.
Reason (R): 8a−8b=8(a−b).

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Answer: (A). The reason correctly proves the closure of multiples of 8 under subtraction.

Q63 1 Mark Ganita Prakash concept

For Assertion–Reason questions use: (A) Both A and R are true and R correctly explains A; (B) Both are true but R is not the correct explanation; (C) A is true but R is false; (D) A is false but R is true.

Assertion (A): The sum of a multiple of 4 and an even non-multiple of 4 is a multiple of 4.
Reason (R): 4p+(4q+2)=4(p+q)+2.

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Answer: (D). The reason is true and shows the sum leaves remainder 2, so the assertion is false.

Q64 1 Mark Ganita Prakash concept

For Assertion–Reason questions use: (A) Both A and R are true and R correctly explains A; (B) Both are true but R is not the correct explanation; (C) A is true but R is false; (D) A is false but R is true.

Assertion (A): If a number is divisible by both 9 and 4, it is divisible by 36.
Reason (R): 9 and 4 are coprime, so LCM(9,4)=36.

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Answer: (A). Divisibility by both implies divisibility by their LCM.

Q65 1 Mark Chapter-based

For Assertion–Reason questions use: (A) Both A and R are true and R correctly explains A; (B) Both are true but R is not the correct explanation; (C) A is true but R is false; (D) A is false but R is true.

Assertion (A): If a number is divisible by both 6 and 4, it must be divisible by 24.
Reason (R): LCM(6,4)=12.

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Answer: (D). The assertion is false; the reason is true and explains why only divisibility by 12 is guaranteed.

Q66 1 Mark Chapter-based

For Assertion–Reason questions use: (A) Both A and R are true and R correctly explains A; (B) Both are true but R is not the correct explanation; (C) A is true but R is false; (D) A is false but R is true.

Assertion (A): A number and the sum of its digits have the same remainder when divided by 9.
Reason (R): Every power of 10 is congruent to 1 modulo 9.

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Answer: (A). The place-value expansion then reduces to the digit sum modulo 9.

Q67 1 Mark Ganita Prakash problem

For Assertion–Reason questions use: (A) Both A and R are true and R correctly explains A; (B) Both are true but R is not the correct explanation; (C) A is true but R is false; (D) A is false but R is true.

Assertion (A): Any rearrangement of the digits of a multiple of 9 remains a multiple of 9.
Reason (R): Rearranging digits does not change their sum.

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Answer: (A). The reason correctly applies the divisibility test for 9.

Q68 1 Mark Chapter-based

For Assertion–Reason questions use: (A) Both A and R are true and R correctly explains A; (B) Both are true but R is not the correct explanation; (C) A is true but R is false; (D) A is false but R is true.

Assertion (A): Every positive multiple of 9 has digital root 9.
Reason (R): Digital root preserves the remainder modulo 9, with remainder 0 represented by 9.

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Answer: (A). This is the defining modular behaviour of digital roots.

Q69 1 Mark Chapter-based

For Assertion–Reason questions use: (A) Both A and R are true and R correctly explains A; (B) Both are true but R is not the correct explanation; (C) A is true but R is false; (D) A is false but R is true.

Assertion (A): The product of three consecutive integers is always divisible by 6.
Reason (R): Among three consecutive integers one is divisible by 3 and at least one is even.

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Answer: (A). These guarantee factors 3 and 2 in the product.

Q70 1 Mark Ganita Prakash problem

For Assertion–Reason questions use: (A) Both A and R are true and R correctly explains A; (B) Both are true but R is not the correct explanation; (C) A is true but R is false; (D) A is false but R is true.

Assertion (A): Doubling a multiple of 11 may destroy divisibility by 11.
Reason (R): If n=11k, then 2n=11(2k).

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Answer: (D). The assertion is false; the reason correctly shows divisibility by 11 is preserved.

Section F — 10 Case Studies

Q71 4 Marks Ganita Prakash theme

Case Study 1 — Parity Lab
Four consecutive integers are 11,12,13,14. Consider all expressions formed by placing + or − between successive numbers.
(a) Find the parity of 11+12+13+14. (b) Find the parity of 11−12+13−14. (c) Explain why all eight expressions have the same parity. (d) State whether this is special to consecutive numbers.

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(a) 11+12+13+14=50, so even.
(b) 11−12+13−14=−2, so even.
(c) Changing any sign changes the value by twice a number, which is even, so parity is unchanged.
(d) No. The same sign-switch argument works for any four integers.

Q72 4 Marks Ganita Prakash problem

Case Study 2 — Remainder Families
A teacher writes the numbers 14,26,38,50,62.
(a) Find their remainder on division by 3. (b) Find their remainder on division by 4. (c) Write a general expression. (d) Explain the role of LCM.

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(a) Each leaves remainder 2 modulo 3.
(b) Each leaves remainder 2 modulo 4.
(c) General form: 12n+2.
(d) Since x−2 must be divisible by both 3 and 4, it must be a multiple of LCM(3,4)=12.

Q73 4 Marks Ganita Prakash problem

Case Study 3 — Digit Sum Detective
A student checks 123,405,8888,93547 and 358095 for divisibility by 9.
(a) Find each digit sum. (b) Which are divisible by 9? (c) Find the remainder of 8888 modulo 9. (d) State the rule.

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(a) Digit sums: 6, 9, 32, 28, 30.
(b) Only 405 has digit sum divisible by 9.
(c) 32 leaves remainder 5 modulo 9, so 8888 leaves remainder 5.
(d) A number is divisible by 9 iff its digit sum is divisible by 9.

Q74 4 Marks Ganita Prakash problem

Case Study 4 — Divisibility by 11
Use the alternating-sum method on 158, 5529 and 90904.
(a) Find the alternating sum for 158. (b) Is 5529 divisible by 11? (c) Is 90904 divisible by 11? (d) State how a nonzero alternating result gives the remainder.

Click to view solution

(a) 8−5+1=4, so 158 leaves remainder 4 modulo 11.
(b) 9−2+5−5=7, so 5529 is not divisible by 11.
(c) 4−0+9−0+9=22, so 90904 is divisible by 11.
(d) Reduce the alternating sum modulo 11 to a remainder from 0 to 10.

Q75 4 Marks Ganita Prakash problem

Case Study 5 — Digital Root Cycle
Starting with 10, repeatedly add 11: 10,21,32,43,54,65,…
(a) List the first nine digital roots. (b) What is the tenth digital root? (c) Why does the pattern cycle? (d) What change modulo 9 does +11 produce?

Click to view solution

(a) Digital roots: 1,3,5,7,9,2,4,6,8.
(b) 109 has digital root 1, restarting the cycle.
(c) Digital roots track residues modulo 9.
(d) Since 11≡2 (mod 9), each step advances by 2 modulo 9.

Q76 4 Marks Ganita Prakash problem

Case Study 6 — Divisibility by 18
The six-digit number is 48a23b and is known to be divisible by 18.
(a) What condition does divisibility by 2 impose on b? (b) Write the digit sum. (c) State the condition for divisibility by 9. (d) List all pairs (a,b).

Click to view solution

(a) b must be even: 0,2,4,6,8.
(b) Digit sum = 17+a+b.
(c) 17+a+b must be a multiple of 9.
(d) Pairs: (1,0),(8,2),(6,4),(4,6),(2,8).

Q77 4 Marks Ganita Prakash problem

Case Study 7 — Divisibility by 44
The number 3p7q8 is divisible by 44.
(a) Which two tests are needed? (b) Which q-values satisfy the last-two-digit test? (c) Form the 11-test equation. (d) List all (p,q).

Click to view solution

(a) Since 44=4×11, use divisibility by 4 and 11.
(b) q∈{0,2,4,6,8}.
(c) 18−(p+q)=0 or a multiple of 11; feasible case gives p+q=7.
(d) (7,0),(5,2),(3,4),(1,6).

Q78 4 Marks Ganita Prakash problem

Case Study 8 — Consecutive Products
A student studies n(n+1), n(n+1)(n+2), and the product of four consecutive integers.
(a) Why is the first always even? (b) Why is the second divisible by 6? (c) What always divides the product of four consecutive integers? (d) What about five?

Click to view solution

(a) One of n,n+1 is even, so the product is divisible by 2.
(b) Among three consecutive integers there is a factor 2 and a factor 3, so divisible by 6.
(c) Four consecutive integers have product divisible by 24.
(d) Five consecutive integers have product divisible by 120.

Q79 4 Marks Ganita Prakash problem

Case Study 9 — Cryptarithm Reasoning
The chapter gives UT×3=PUT and AB×5=BC.
(a) State U,T,P for the first. (b) Verify numerically. (c) State A,B,C for the second. (d) Verify numerically.

Click to view solution

(a) U=5,T=0,P=1.
(b) 50×3=150.
(c) A=1,B=9,C=5.
(d) 19×5=95.

Q80 4 Marks Ganita Prakash theme

Case Study 10 — Always, Sometimes, Never
Classify the following:
(a) sum of two even numbers is a multiple of 3; (b) sum of a multiple of 6 and a multiple of 9 is a multiple of 3; (c) odd+even is a multiple of 6; (d) doubling a multiple of 11 remains a multiple of 11.

Click to view solution

(a) Sometimes true: 2+4=6, but 2+6=8.
(b) Always true: 6x+9y=3(2x+3y).
(c) Never true: odd+even is odd, but every multiple of 6 is even.
(d) Always true: 2(11k)=11(2k).

Ganita Prakash Problems

This final section organises the problems and solutions supplied with the uploaded Ganita Prakash chapter. It is separate from the 80-question bank above.

Page 122 — Figure it Out 1: Four consecutive numbers

The PDF answer gives the four consecutive numbers as 7, 8, 9, 10. Algebraically, n+(n+1)+(n+2)+(n+3)=34 gives n=7.

Page 122 — Figure it Out 2: Five consecutive numbers

If p is the greatest of five consecutive integers, the other four are p−1, p−2, p−3, p−4.

Page 122 — Figure it Out 3(i): Sum of two even numbers is a multiple of 3

Sometimes true. The PDF gives examples 2+4=6 and 4+8=12, but non-examples 2+6=8 and 6+8=14.

Page 122 — Figure it Out 3(ii): Not divisible by 18 ⇒ not divisible by 9

Sometimes true. The PDF uses 30 as a number divisible by neither 18 nor 9, and 27 as a number not divisible by 18 but divisible by 9.

Page 122 — Figure it Out 3(iii): Two non-multiples of 6

Sometimes true. 9 and 11 are not divisible by 6 and sum to 20, also not divisible by 6; but 8 and 10 are not divisible by 6 while their sum 18 is divisible by 6.

Page 122 — Figure it Out 3(iv): Multiple of 6 + multiple of 9

Always true. 6x+9y=3(2x+3y), so the sum is always a multiple of 3.

Page 122 — Figure it Out 3(v): Multiple of 6 + multiple of 3 is a multiple of 9

Sometimes true. 18+9=27 is a multiple of 9, whereas 12+9=21 is not.

Page 122 — Figure it Out 4: Same remainder 2 modulo 3 and 4

The PDF derives x=12n+2. Examples are 14, 26, 38, … because x−2 must be divisible by LCM(3,4)=12.

Page 122 — Figure it Out 5: Pebbles riddle

The PDF identifies the number as 91. The conditions give candidates 31,61,91 below 100, and 91 is divisible by 7.

Page 122 — Figure it Out 6: Three numbers each ≡2 (mod 6)

The claim is true. Writing the three numbers as 6a+2, 6b+2, 6c+2 gives sum 6(a+b+c+1).

Page 123 — Figure it Out 7: Remainders modulo 7

Given 4779=7p+5 and 661=7q+3:
(i) 4779+661 leaves remainder 1.
(ii) 4779−661 leaves remainder 2.

Page 123 — Figure it Out 8: Remainders one below 3,4,5

The smallest number is 59, because N+1 must be a multiple of LCM(3,4,5)=60.

Page 126 — Divisibility by 9, Q1

Among 123, 405, 8888, 93547 and 358095, the PDF states that only 405 is divisible by 9.

Page 126 — Divisibility by 9, Q2

The PDF gives 288 as the smallest multiple of 9 with no odd digits. Its digits sum to 18.

Page 126 — Divisibility by 9, Q3

The multiple of 9 closest to 6000 is 6003.

Page 126 — Divisibility by 9, Q4

There are 11 multiples of 9 strictly between 4300 and 4400.

Page 128 — Divisibility by 11

If the alternating difference is 0 or a multiple of 11, the number is divisible by 11. The PDF gives: 158 → remainder 4; 841 → 5; 481 → 8; 5529 → 7; 90904 and 857076 are divisible by 11.

Page 129 — Divisibility table

The PDF table records divisibility by 2,3,4,5,6,8,9,10,11 for 128, 990, 1586, 275, 6686, 639210, 429714, 2856, 3060 and 406839. For example, 990 is divisible by 2,3,5,6,9,10,11 but not by 4 or 8.

Page 130 — Digital root and division by 9

The PDF states that the digital root gives the remainder on division by 9, with an exactly divisible positive number having digital root 9.

Page 130 — Digital roots between 600 and 700

The PDF lists:
Digital root 5: 608,617,626,635,644,653,662,671,680,689,698.
Digital root 7: 601,610,619,628,637,646,655,664,673,682,691.
Digital root 3: 606,615,624,633,642,651,660,669,678,687,696.

Page 130 — Digital-root cycles

Consecutive multiples of 3: 3,6,9 repeating.
Multiples of 4: 4,8,3,7,2,6,1,5,9 repeating.
Multiples of 6: 6,3,9 repeating.

Page 131 — Figure it Out 1: Digital root after adding 10

The PDF answer is 6 when the original 8-digit number has digital root 5.

Page 131 — Figure it Out 2: Repeatedly adding 11

Starting with 10 gives 10,21,32,43,54,… with digital roots 1,3,5,7,9,2,4,6,8,1,….

Page 131 — Figure it Out 3: 9a+36b+13

The PDF rewrites it as 9(a+4b+1)+4, so the digital root is 4.

Page 131 — Figure it Out 4: Digital root relations

The PDF says there is no consistent relation between parity and digital root. For division by 3, roots 1,4,7 correspond to remainder 1; 2,5,8 to remainder 2; 3,6,9 to remainder 0. For division by 9, root 9 corresponds to remainder 0; roots below 9 equal the remainder.

Page 131 — Addition cryptarithms

Solutions supplied in the PDF:
(i) A=7, B=9.
(ii) A=2, B=5.
(iii) N=1, O=3, P=9.
(iv) Q=8, R=5, P=2.

Page 132 — Multiplication cryptarithms

Solutions supplied in the PDF:
UT×3=PUT: U=5,T=0,P=1.
AB×5=BC: A=1,B=9,C=5.
L2N×2=2NP: L=1,N=5,P=0.
XY×4=ZX: X=2,Y=3,Z=9.
PP×QQ=PRP: P=2,Q=1,R=4.
JK×6=KKK: J=7,K=4.

Page 132 — Figure it Out 1: 31z5 divisible by 9

The PDF gives z=0 or 9, because the digit sum is 9+z.

Page 132 — Figure it Out 2: Snehal’s claim

Let a=12n+8 and b=12m−4. Then a+b=12(n+m)+4. The PDF concludes the sum is not always a multiple of 8.

Page 132 — Figure it Out 3: Sum of two multiples of 3

For 3m and 3n, the sum is 3(m+n). It is a multiple of 6 exactly when m+n is even.

Page 132 — Figure it Out 4: Reversing digits of a multiple of 9

The conjecture is true. Reversing—or any other rearrangement of—the digits preserves the digit sum, so divisibility by 9 is preserved.

Page 132 — Figure it Out 5: 48a23b divisible by 18

Possible pairs are (1,0), (8,2), (6,4), (4,6), (2,8).

Page 133 — Figure it Out 6: 3p7q8 divisible by 44

Possible pairs are (p,q)=(7,0),(5,2),(3,4),(1,6).

Page 133 — Figure it Out 7: Consecutive 2,3,4 divisibility pattern

One triple is 2,3,4. The pattern repeats every 12, so the next is 14,15,16. In general: 12k+2, 12k+3, 12k+4.

Page 133 — Figure it Out 8: Multiples of 36

The PDF starts the list with 45036, 45072, 45108, … and asks students to continue. These satisfy the tests for both 4 and 9.

Page 133 — Figure it Out 9: Five consecutive even numbers

If the middle term is 5p, the other terms are 5p−4, 5p−2, 5p+2, 5p+4.

Page 133 — Figure it Out 10: Six-digit number and its reversal

The PDF explains that the original number must end in 5 and begin with an even nonzero digit, while the digit sum must be divisible by 3. It gives examples beginning with 200025, 200055, 200085, 202005, ….

Page 133 — Figure it Out 11: Doubling multiples of 11

Deepak’s conjecture is false. If n=11k, then 2n=22k=11(2k), so every doubled multiple of 11 is still divisible by 11.

Page 133 — Figure it Out 12: Always/Sometimes/Never

The PDF classifications are:
(i) product of a multiple of 6 and a multiple of 3 is a multiple of 9 — Always true.
(ii) sum of three consecutive even numbers divisible by 6 — Always true.
(iii) if abcdef is a multiple of 6, badcef is a multiple of 6 — Always true.
(iv) 8(7b−3)−4(11b+1) is a multiple of 12 — Never true.

Page 133 — Figure it Out 13: Sum of any three numbers and divisibility by 3

The PDF analyses residues 0,1,2. The sum is divisible by 3 when the three remainders add to a multiple of 3, e.g. 0+1+2 or three equal remainders such as 0+0+0, 1+1+1 or 2+2+2.

Page 133 — Figure it Out 14: Consecutive products

The PDF states: product of 2 consecutive integers is divisible by 2; 3 consecutive by 6; 4 consecutive by 24; 5 consecutive by 120.

Page 133 — Figure it Out 15: Cryptarithms

The PDF solutions are:
(i) EF×E=GGG: E=3,F=7,G=1 (37×3=111).
(ii) WOW×5=MEOW: W=5,O=7,M=2,E=8 (575×5=2875).

Page 133 — Figure it Out 16: Venn diagram

The correct diagram is (iv): multiples of 32 are contained within multiples of 8, which are contained within multiples of 4.