NCERT Solutions for Class 8 Maths Chapter 5 Number Play (PDF) – 2027

No. The powers of 2 — 1, 2, 4, 8, 16, 32, … — cannot be written as a sum of two or more consecutive natural numbers. Every other natural number can.

Check the smallest cases:

Numbers that have three or more odd factors. In fact the number of ways (using two or more consecutive natural numbers) is exactly one less than the number of odd factors.

Anshu’s first claim is easy to justify, and the answer to his question is no — the even numbers 2, 4, 8, 16, 32, … (the powers of 2) cannot be written this way.

Yes, once negative numbers are allowed. Take any run that is symmetric about 0 — the positive and negative parts cancel.

Eight possibilities. There are 3 gaps between the four numbers and each gap takes either sign, so the count is 2 × 2 × 2 = 2³ = 8.

The tree in the book lists them in exactly this order: branch on the sign before 4, then before 5, then before 6.

The eight values for 3, 4, 5, 6 are 18, 6, 8, – 4, 10, – 2, 0, – 12 — and the interesting thing is that every one of them is even.

Notice also that no value is odd and no value is repeated.

Taking 5, 6, 7, 8:

Observation: again all eight values are even. And three of them — 0, – 2 and – 4 — are exactly the same as before.

Taking 7, 8, 9, 10:

The three constant answers appear once more. Writing the numbers as n, n + 1, n + 2, n + 3 shows why they must:

Yes. Algebra settles it without testing more sets.

• 1 ± 1 ± 1 ± 1 is one of 4, 2, 0, – 2 — always even. So the first part is an even multiple of n.
• 0 ± 1 ± 2 ± 3 starts from 1 + 2 + 3 = 6, which is even, and changing any sign alters it by twice that number — again an even change. So the second part is always even too.

All eight expressions have the same parity — and it is the parity of a + b + c + d.

Yes. The key is that switching one sign changes the value by an even number, so it can never change the parity.

If two numbers differ by an even number, they cannot have different parities — either both are even or both are odd.

Change – c to + c:

No — it holds for any count of numbers. Nothing in the argument used the number four.

Four of them are even: 43 + 37, 672 – 348, 4 × 347 × 3 and 543 – 479.

Five of the nine are always even: 2a + 2b, 4m + 2n, 2u – 4v, 13k – 5k and 4k × 3j. Each of them has 2 as a factor of the whole expression.

Handle each one the way the book handles 4m + 2q and x² + 2 — either pull out the factor 2, or produce a non-example.

Any expression with 2 as a factor of the whole thing will do:

Sometimes yes, sometimes no — and the pattern is exact.

The rule: divide each even number by 4 — the remainder is either 0 or 2. The sum is a multiple of 4 exactly when the two remainders are the same.

Exactly when both are multiples of 4, or neither is. There are three cases to examine, and algebra settles all three:

Yes, it is the same problem one level up. Compare the two tables:

The sum is never a multiple of 4 — it always leaves a remainder of 2.

Yes, it is exactly similar. Adding an even and an odd number can never give an even number; adding a multiple of 4 and a non-multiple of 4 can never give a multiple of 4. In both cases the two remainders are different (0 and 1 there, 0 and 2 here), so the leftover cannot be cleared.

Explanation. A multiple of 4 can be written as 4p, and an even number that is not a multiple of 4 as 4q + 2. Adding them gives 4(p + q) + 2, so the sum is a multiple of 4 plus a leftover 2. It is even, but it is never a multiple of 4.

Examples.

Always true for addition, and always true for subtraction as well.

Sometimes true. The word that breaks it is any.

Always true.

Take 14 = 7 × 2 (so j = 2). Its multiples:

Always true.

Take 24, which is divisible by 12. Its factors include every factor of 12: 1, 2, 3, 4, 6, 8, 12, 24.

Sometimes true. 42 is divisible by 7 and by 14, but not by 28.

Always true.

If the number is divisible by 9, its prime factorisation contains 3 × 3. If it is also divisible by 4, the factorisation contains 2 × 2. Since these are different primes, both blocks sit in the factorisation together, so 3 × 3 × 2 × 2 = 36 divides the number.

Sometimes true. The counterexample is small: 12.

Never true. The sum of an odd and an even number is odd, while every multiple of 6 is even.

The left side is even; the right side is 1 less than an even number, so it is odd. An even number can never equal an odd number, so no values of n, m, j exist.

Not for this statement. 2n and 2n + 1 are always consecutive numbers (like 4 and 5), so that choice would prove the claim only for an even number followed by the next odd number. The statement is about any even number and any odd number (like 4 and 11), so we need two different letters: 2n and 2m + 1.

3, 8, 13, 18, 23, 28, 33, 38, … — each one is 3 more than a multiple of 5.

The gap between consecutive numbers in the list is always 5, because moving to the next multiple of 5 keeps the leftover 3 unchanged.

(iv) 5k + 3 and (v) 5k – 2. They look different but describe exactly the same set of numbers.

Infinitely many. Any expression 5k + c works, provided c itself leaves a remainder of 3 on division by 5.

The numbers are 7, 8, 9 and 10.

Check: 7 + 8 + 9 + 10 = 34 ✓

The other four are p – 1, p – 2, p – 3 and p – 4.

(i) Sometimes true. 2m + 2n = 2(m + n) is a multiple of 3 only when m + n is. Example: 2 + 4 = 6 ✓; non-example: 2 + 6 = 8 ✗.

(ii) Sometimes true. Example: 20 is divisible by neither 18 nor 9 ✓; non-example: 27 is not divisible by 18, but 27 = 9 × 3 ✗ (18k = 9 × 2k, so only every second multiple of 9 is a multiple of 18).

(iii) Sometimes true. (6a + r) + (6b + s) = 6(a + b) + (r + s), with remainders r, s from 1 to 5; the sum is divisible by 6 only when r + s = 6. Example: 8 + 9 = 17 ✓; non-example: 8 + 10 = 18 ✗.

(iv) Always true. 6x + 9y = 3(2x + 3y). Example: 12 + 9 = 21 = 3 × 7.

(v) Sometimes true. 6x + 3y = 3(2x + y) is a multiple of 9 only when 2x + y is a multiple of 3. Example: 18 + 9 = 27 ✓; non-example: 12 + 9 = 21 ✗.

Such numbers are 2, 14, 26, 38, 50, 62, … and they are exactly the numbers of the form 12n + 2.

There are 91 pebbles.

First turn each line into a condition:

Yes, Tathagat’s claim is true.

(i) remainder 1   (ii) remainder 2.

Algebraically. Write each number as a multiple of 7 plus its remainder:

Visually. Think of each number as counters arranged in rows of 7, with the loose ones at the bottom.

The smallest such number is 59.

Look at the three conditions side by side: each remainder is exactly 1 less than its divisor.

Write the number as a sum of place values: N = … + 1000d + 100c + 10b + a.

All of them.

Yes. If every digit is 0 or 9, then in the expanded form every single term is a multiple of 9.

No — 10 leaves a remainder of 1, because 10 = 9 + 1.

So for any multiple of 10 the remainder is the number of tens.

The remainder is the number of hundreds.

The remainder is 4.

Check: 427 = 9 × 47 + 4 = 423 + 4 ✓

Yes — for every place value, however large. Every place value is 1 more than a multiple of 9 (10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, …).

So each digit contributes exactly itself to the remainder. For 7309: 7309 = (a multiple of 9) + (7 + 3 + 0 + 9) = (a multiple of 9) + 19.

Now shrink 19 the same way: 1 + 9 = 10, then 1 + 0 = 1. So 7309 is 1 more than a multiple of 9, and 7309 ÷ 9 leaves remainder 1.

All four are correct.

Everything follows from one fact established on this page:

Only (ii) 405 is divisible by 9.

The smallest one is 288.

Two digits? The largest possible digit sum is 8 + 8 = 16, which is less than 18. Impossible.

Three digits? Keep the hundreds digit as small as possible, then the tens:

The closest multiple of 9 is 6003.

6003 is only 3 away, so it is the closer of the two.

There are 11 of them.

The eleven numbers are 4302, 4311, 4320, 4329, 4338, 4347, 4356, 4365, 4374, 4383, 4392.

Every power of 10 leaves a remainder of 1 when divided by 3.

Yes, 462 is divisible by 11.

Use the alternating behaviour of the place values: 1 and 100 are 1 more than a multiple of 11, while 10 is 1 less.

Nothing is left over, so 462 is a multiple of 11. Check: 462 = 11 × 42 ✓

Here is the shortcut, in the two forms the chapter uses.

Form 1 — excess and short.

1. Add the digits sitting in the places 1, 100, 10000, … (these are 1 more than a multiple of 11). Call this the excess.
2. Add the digits sitting in the places 10, 1000, 100000, … (these are 1 less than a multiple of 11). Call this the short.
3. Compute excess – short. If it is 0 or a multiple of 11, the number is divisible by 11; otherwise it tells you how far off you are.

The remainder is 0 — the number is exactly divisible by 11.

Now suppose the difference had come out as 11, or 22, or – 11. That leftover is itself a whole number of 11s, so it joins the multiple of 11 already built up and nothing at all remains.

Attach alternating ‘+’ and ‘–’ signs starting from the units digit.

It is the same method, written more compactly.

The quick way — do only four checks per row and read off the rest:

• 2 — units digit even. 5 — units digit 0 or 5. 10 — units digit 0 (so 10 = 2 and 5 together).
• 3 and 9 — one digit sum serves both. If the digit sum is a multiple of 9 it is automatically a multiple of 3.
• 6 — no new work: it is just 2 and 3.
• 4 — last two digits; 8 — last three digits.
• 11 — alternating sum of the digits from the units place.

Check divisibility by 2 and by 3 — both must hold.

Yes, it works every time.

Notice that 38 and 64 pass the test for 2 but fail for 3, and 225 does the opposite. Both tests must be passed.

No, it does not work. The smallest counterexample is 12.

The chapter’s replacement test is to check divisibility by 3 and by 8 instead.

Write everything in prime factors:

Using 3 and 8. Divisibility by 8 puts 2³ into the number’s prime factorisation; divisibility by 3 puts a 3 there. These use different primes, so both blocks sit side by side:

Using 4 and 6. Divisibility by 4 puts 2² in; divisibility by 6 puts 2 × 3 in. But the 2 inside 6 can be the same 2 that is already inside 4 — the two demands overlap.

The digital root of a number is its remainder on division by 9 — except that a multiple of 9 has digital root 9 rather than 0.

Every number here is 6ab, so the digit sum is 6 + a + b. Work out what a + b must be in each case.

(i) Digital root 5 → 6 + a + b must be 5, 14 or 23 → a + b = 8 or 17

(ii) Digital root 7 → 6 + a + b must be 7 or 16 → a + b = 1 or 10

(iii) Digital root 3 → 6 + a + b must be 12 or 21 → a + b = 6 or 15

Take 25 to 36:

Observation: the digital roots go up by 1 each time, and after 9 they start again at 1. The pattern repeats every 9 numbers — that is why the 12 roots show the first three of them twice.

(i) The roots cycle 3, 6, 9 — only three values, repeating every 3 multiples.

(ii) The roots cycle 4, 8, 3, 7, 2, 6, 1, 5, 9 — all nine values appear before repeating.

(iii) The roots cycle 6, 3, 9 — again only three values.

The roots cycle 7, 4, 1.

What you notice: only three of the nine possible roots ever appear, and they are exactly the roots that leave remainder 1 on division by 3.

The number is 11,11,11,111 — nine 1s in a row.

Its name in the Indian system: eleven crore eleven lakh eleven thousand one hundred eleven.

Read the clues one by one:

The new digital root is 6.

Start at 10 and keep adding 11:

Observations.

• The digital root goes up by 2 at every step, wrapping round after 9.
• All nine roots appear, and the whole pattern repeats after 9 terms.
• Reading the roots in order gives 1, 3, 5, 7, 9, 2, 4, 6, 8 — the odd roots first, then the even ones.

The digital root is 4, whatever whole numbers a and b are.

The bracket is multiplied by 9, so that part is a multiple of 9 and contributes nothing to the leftover. What remains is 4.

(i) Parity and digital root — there is no relation.

All four combinations occur, so knowing the parity tells you nothing about the digital root and vice versa.

(ii) Digital root and remainders — there is a clean relation both times.

Division by 9: the remainder equals the digital root, except that a digital root of 9 means remainder 0.

Division by 3: the remainder is what the digital root itself leaves on division by 3.

(i) 71 + 19 = 90 (A = 7, B = 9). 1 + B ends in 0 → B = 9, carry 1; A + 1 + 1 = 9 → A = 7.

(ii) 25 + 37 = 62 (A = 2, B = 5). The tens need a carry (else B < 0), so A + 3 + 1 = 6 → A = 2; B + 7 = 12 → B = 5.

(iii) Three solutions: 17 × 3 = 51, 24 × 3 = 72, 31 × 3 = 93 (e.g. O = 3, N = 1, P = 9). 3 × ON < 100 → O = 1, 2 or 3, and 3 × N must end in O.

(iv) 85 + 85 + 85 = 255 (Q = 8, R = 5, P = 2). 3R ends in R → R = 5 (R = 0 fails); 3 × Q5 = P55 → Q = 8.

No, PQ cannot be 13 — and 12 is the only value that works.

For the product to stay at two digits we need PQ × 8 ≤ 99, that is PQ ≤ 12. So only 10, 11 and 12 are candidates:

The solution is 24 × 4 = 96, giving G = 2, H = 4, K = 6.

The letter to watch is H: it is both the units digit of the two-digit number and the multiplier. So the multiplier must equal the units digit.

Y must be even and less than 7, so the only possibilities are Y = 0, 2, 4 or 6. Testing them leaves exactly one solution.

• Y is even. The product RAY ends in Y, and 6 × (anything) is even. So Y cannot be 1, 3, 5, 7 or 9.
• Y is less than 7. If Y = 7, then BYE is at least 170, and 170 × 6 = 1020 — a 4-digit product. So 7, 8 and 9 are out.

Now try each surviving value, remembering B = 1 and that 6 × E must end in Y:

(i) 50 × 3 = 150 (U = 5, T = 0, P = 1). 3 × UT = 100P + UT, so UT = 50P → P = 1.

(ii) 19 × 5 = 95 (A = 1, B = 9, C = 5). AB ≤ 19, so A = 1; 50 + 5B = 10B + C → C = 50 – 5B → B = 9.

(iii) 125 × 2 = 250 (L = 1, N = 5, P = 0) or 124 × 2 = 248 (L = 1, N = 4, P = 8). L = 1, and 240 + 2N = 200 + 10N + P → 8N + P = 40.

(iv) 23 × 4 = 92 (X = 2, Y = 3, Z = 9). XY ≤ 24, and X is the last digit of 4 × Y, so X is even → X = 2; 4Y ends in 2 → Y = 3.

(v) 22 × 11 = 242 (P = 2, Q = 1, R = 4); also 33 × 11 = 363 and 44 × 11 = 484. PP × QQ = 121 × P × Q.

(vi) 74 × 6 = 444 (J = 7, K = 4). 60J + 6K = 111K → 4J = 7K → K = 4, J = 7.

z = 0 or z = 9, giving 3105 and 3195.

Snehal’s claim is false. The sum is sometimes a multiple of 8 and sometimes not.

So the sum is always a multiple of 4, but for it to be a multiple of 8 the bracket 3(n + m) + 1 must be even — that is, n + m must be odd.

The sum of two multiples of 3 is a multiple of 6 exactly when the two numbers have the same parity — both even or both odd.

(i) Her conjecture is true for every multiple of 9.

(ii) Yes — any shuffle of the digits works, not just the reversal.

All six are multiples of 9, for the same reason: rearranging digits never changes their sum.

The pairs are (a, b) = (1, 0), (8, 2), (6, 4), (4, 6) and (2, 8).

Divisible by 2: the units digit b must be even → b = 0, 2, 4, 6 or 8.

Divisible by 9: the digit sum must be a multiple of 9.

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

Divisible by 4: the last two digits q8 must form a multiple of 4 → 08, 28, 48, 68, 88, so q = 0, 2, 4, 6 or 8.

One such set is 2, 3, 4. There are infinitely many, and they occur once in every 12 numbers.

Five of them: 45036, 45072, 45108, 45144, 45180.

The approach. Find one multiple of 36 in the range, then keep adding 36 — every step lands on the next one.

The other four are 5p – 4, 5p – 2, 5p + 2 and 5p + 4.

Consecutive even numbers differ by 2, not 1 — that is the only thing to be careful about here.

One answer: 210045.

Deepak’s conjecture is false. Every multiple of 11 stays a multiple of 11 when doubled — there are no exceptions.

11 is still a factor, so 2N is a multiple of 11 no matter what k is.

(i) Always true. 6m × 3n = 18mn = 9 × (2mn). Example: 12 × 9 = 108 = 9 × 12.

(ii) Always true. 2n + (2n + 2) + (2n + 4) = 6n + 6 = 6(n + 1). Example: 4 + 6 + 8 = 18 = 6 × 3.

(iii) Always true. badcef has the same units digit f, so it is still even, and the same digits, so the same digit sum — still a multiple of 3. Example: 123456 = 6 × 20576 and 214356 = 6 × 35726.

(iv) Never true. 8(7b – 3) – 4(11b + 1) = 56b – 24 – 44b – 4 = 12b – 28 = 12(b – 2) – 4, which is always 4 less than a multiple of 12.

Divide each number by 3 and keep only its remainder — 0, 1 or 2. The sum is divisible by 3 exactly when the three remainders add to 0, 3 or 6.

Since each remainder is 0, 1 or 2, their total lies between 0 and 6. Only 0, 3 and 6 work, and that leaves exactly two shapes of answer:

Generalisation: the sum of three numbers is divisible by 3 when their remainders on division by 3 are all equal or all different — and in no other case.

Two consecutive integers — always a multiple of 2: one of n and n + 1 is even. Example: 8 × 9 = 72.

Three — always a multiple of 6: at least one of them is even and exactly one is a multiple of 3, and 2 × 3 = 6. Example: 4 × 5 × 6 = 120 = 6 × 20.

Four — always a multiple of 24: two of them are even and one of these two is a multiple of 4 (together a factor 8), and one is a multiple of 3; 8 × 3 = 24. Example: 5 × 6 × 7 × 8 = 1680 = 24 × 70.

Five — always a multiple of 120: the first four already give 24, and one of the five is a multiple of 5; 24 × 5 = 120. Example: 2 × 3 × 4 × 5 × 6 = 720 = 120 × 6.

(i) E = 3, F = 7, G = 1: 37 × 3 = 111. GGG = 111 × G = 3 × 37 × G, so EF × E is one of 111, 222, …, 999; the multiplier E must equal the tens digit of EF.

(ii) W = 5, O = 7, M = 2, E = 8: 575 × 5 = 2875. 5 × W ends in W and W ≠ 0, so W = 5. Then (505 + 10O) × 5 = 1000M + 100E + 10O + 5 gives 2520 + 40O = 1000M + 100E, so 2520 + 40O must end in 00 → O = 2 or 7. O = 2 gives 2600 (M = 2 = O ✗); O = 7 gives 2800 → M = 2, E = 8.

Diagram (iv) — the one that nests multiples of 32 inside multiples of 8 inside multiples of 4.