NCERT Solutions for Class 6 Maths Chapter 5 Prime Time (PDF) – 2027

These numbers are called common multiples of 3 and 5.

The players say ‘idli-vada’ exactly at the numbers that are in the 3 table and in the 5 table.

The numbers that appear in both lists are

‘Idli-vada’ is said at the common multiples of 3 and 5, that is, at the multiples of 15.

You do not have to count them one by one:

Answer: at the number 150.

Just count how many multiples of 3, of 5 and of 15 there are up to 90.

The 6 ‘idli-vada’ numbers are 15, 30, 45, 60, 75 and 90.

900 is exactly 10 times 90, so every count becomes 10 times bigger.

The 10th ‘idli-vada’ is still at 150, but now the game goes on till the 60th one, at 900.

Yes. Fig. 5.1 is the ‘idli-vada’ game played till 30: the left circle holds the ‘idli’ numbers (multiples of 3), the right circle the ‘vada’ numbers (multiples of 5), and the overlap the ‘idli-vada’ numbers 15 and 30 (the common multiples).

Playing till 60 gives the same picture with more numbers in it:

For each pair, first find the first common multiple. Then every next common multiple is obtained by adding that number again.

Answer: 8.

Nobody said just ‘vada’. That means every multiple of the other number was also a multiple of 4 — so it always turned into an ‘idli-vada’. This happens only when the other number is itself a multiple of 4.

Jump sizes 1, 3, 5 and 15. A jump size reaches a number only if it is a factor of that number, so it must be a common factor of 15 and 30.

The numbers in both lists are

The shaded numbers in the table (31 to 70) are

Yes — every one of them is in the 3 table.

All the shaded numbers are multiples of 3. They also step up by 3 each time — 33, 36, 39, … — so they sit in a slanting pattern in the table.

The circled numbers are

Each one goes up by 4 from the previous one, and each is in the 4 table.

All the circled numbers are multiples of 4.

Only three numbers wear both marks:

They are common multiples of 3 and 4.

Write the 40 table and pick the numbers that fall in between.

Between 310 and 410 we get

a. The number is 35.

“One of my factors is 7” means the number is a multiple of 7. The multiples of 7 below 40 are:

b. The number is 45.

Having both 3 and 5 as factors means the number is a multiple of 15. Below 100 these are:

The perfect number is 6.

List all the factors of each number and pick out the ones that appear everywhere.

Write the 25 table and cross out the numbers that also appear in the 50 table.

Three such numbers: 25, 75 and 125. (175, 225, 275 … also work.)

The first ‘idli-vada’ happens at the first common multiple of the two numbers. So we need two numbers below 10 whose first common multiple is more than 50.

Answer: the two numbers could be 7 and 8, or 7 and 9, or 8 and 9.

Jumpy needs a common factor of 28 and 70.

The numbers in both lists are

Answer: jump sizes 1, 2, 7 and 14.

The overlap shows 24, 48 and 72. These go up by 24 each time, so the first common multiple of the two numbers is 24.

So we need a pair whose first common multiple is 24. A neat choice is 8 and 12:

We need the smallest number divisible by 1, 2, 3, 4, 5, 6, 8, 9 and 10. Take each prime as many times as any one of these numbers needs it.

This is the previous answer with the missing 7 put back.

1. Guna’s 12 figs — 12 = 1 × 12 = 2 × 6 = 3 × 4, and each can be turned around:

2. Anshu’s 7 figs — 7 can only be split as 1 × 7:

In every arrangement, rows × columns = 12. So the number of rows and the number of columns are always factors of 12.

Reading the top row we get 12, 6, 4, 3, 2, 1 — exactly the list of factors of 12.

Test each of the ten numbers from 21 to 30.

2 + 8 = 10, and there are exactly 10 numbers from 21 to 30. ✔

Yes. Write 1 to 100 in a grid and follow the five steps — cross out 1, then circle 2 and cross out its later multiples, circle 3 and cross out its later multiples, then 5, then 7. After that everything left is already circled.

The 25 prime numbers below 100 are

The sieve works because of one simple idea: a composite number always has a smaller prime as a factor, so it is bound to get crossed out at some step.

No. 2 is the only even prime number.

Write the primes and the gaps between them:

Smallest difference = 1, between 2 and 3.

Largest difference = 8, between 89 and 97.

No, the primes are not spread out equally. Counting them row by row:

Least: the decade 91 – 100, with only one prime (97).

Most: the decades 1 – 10 and 11 – 20, with four primes each.

For each number, look for a factor other than 1 and itself.

Answer: 23 and 37 are prime. 51 and 26 are composite.

The primes below 20 are 2, 3, 5, 7, 11, 13, 17, 19. Pick pairs whose sum ends in 0 or 5.

Three such pairs: (2, 3), (3, 7) and (2, 13).

Take each two-digit prime, reverse its digits, and check whether the reversed number is also prime.

Answer: the pairs are (13, 31), (17, 71), (37, 73) and (79, 97).

Answer: 90, 91, 92, 93, 94, 95, 96.

Look for the biggest gap between two primes below 100. It comes between 89 and 97, and all the seven numbers in between are composite.

Run down the list of primes and note every pair that differs by 2.

The book already gives (3, 5) and (17, 19), so the other twin primes are

There are 8 twin-prime pairs below 100 in all.

Answer: 105.

Find the prime factorisation of each and count the different primes in it.

Answer: none — zero such primes.

The digits 2, 4 and 5 can be arranged in six ways:

Yes, there are many. Double the prime and add 1, then check the answer.

Five examples: 2 → 5, 5 → 11, 11 → 23, 23 → 47 and 29 → 59. (41 → 83 is a sixth.)

Remember the new rule — a jump size of 1 is not allowed. So a pair is safe only if the two numbers have no common factor other than 1.

Safe pairs: (4, 15) and (18, 29).

Co-prime pairs: a (18 and 35), b (15 and 37) and d (17 and 69).

The rule Anshu spotted:

• If the two numbers are co-prime, the first common multiple is exactly their product.
• If they share a common factor, the first common multiple is less than the product.

Read each picture by counting the pegs on the circle and the pegs the thread actually touches.

Yes, exactly. The thread reaches every peg if and only if the number of pegs and the thread-gap are co-prime.

First decide, for each pair, how many pegs the thread will touch. Then draw.

Anshu’s mistake was to think that one way of splitting a number shows all its factors. It does not.

The number 7 is a factor of both, but Anshu’s two splittings simply did not show it. Guna’s splittings (56 = 7 × 8 and 63 = 9 × 7) happened to show it.

No — not from those splittings alone. Another way of splitting might show a common factor, as happened with Anshu. So we break both numbers down to primes:

The primes of 80 are 2 and 5; the primes of 63 are 3 and 7. Nothing is shared, and now there is nowhere left to hide.

So yes — 80 and 63 really are co-prime, but the prime factorisation is what proves it.

The picture is a box built out of small cubes, 2 cubes tall, 3 cubes wide and 5 cubes deep. Count the cubes in whichever order you like — the total never changes.

Put the two little factorisations side by side.

Writing the primes in increasing order:

The counts match exactly. So when two numbers are multiplied, the prime factors of the product are simply the prime factors of the two numbers put together.

Keep dividing by the smallest prime that works, until only primes are left.

Just multiply the primes as many times as they occur.

Answer: 198.

Break 1955 down step by step. It ends in 5, so start with 5.

All three — 5, 17 and 23 — are prime and all are less than 30. ✔

To keep the answer as small as possible, use the smallest primes and use each of them only once.

The first number is divisible by the second only if the whole prime factorisation of the second sits inside that of the first.

Are they co-prime? No. They share the primes 3 and 7, so 3, 7 and 21 are all common factors.

Does one divide the other? No, neither divides the other.

Yes, Guna is right — as long as the two primes are different.

125 is not a multiple of 10, and it will never appear in that sequence.

Why not: every multiple of 10 ends in 0, but 125 ends in 5.

Is 8560 divisible by 10? Yes.

Yes, we fully agree. This works both ways round.

• Every multiple of 10 ends in 0 — because multiplying by 10 shifts every digit one place to the left and drops a 0 into the units place. 7 × 10 = 70, 43 × 10 = 430.
• Every number ending in 0 is a multiple of 10 — because you can rub off the last 0 and read the rest as the quotient. 2560 = 256 × 10.

List the multiples of 5 and look at the last digit:

The last digit is always 0 or 5, turn and turn about.

Largest number less than 399 divisible by 5 = 395.

Is 8560 divisible by 5? Yes.

Yes, we agree.

Look at the last digits of the multiples of 2:

The last digit is always one of 0, 2, 4, 6, 8, repeating in a cycle of five.

Is 682 divisible by 2? Yes — and we do not need long division.

Is 8560 divisible by 2? Yes, because it ends in 0.

Yes, we agree — these are exactly the even numbers.

Multiples of 2 between 399 and 411:

Multiples of 4 from 1 to 200:

Pattern: after 100 the endings repeat — 104, 108, 112, … end in 04, 08, 12, … just like 4, 8, 12, …. So a multiple of 4 is decided by its last two digits.

What we observe: in all three stretches the same two endings turn up — 32 and 36. The digits in front make no difference at all. So only the last two digits decide divisibility by 4.

Is 8536 divisible by 4? Yes.

Yes — all three statements are true.

What we observe: the same two endings — 128 and 136 — appear every time. The thousands digit changes nothing. So only the last three digits decide divisibility by 8.

Can the last two digits be used? No. 128 and 228 both end in 28, but 128 = 8 × 16 while 228 ÷ 8 leaves remainder 4.

Change the last two digits so that the last three digits still make a multiple of 8. Any of these works:

Yes — all three statements are true.

Factors of 8560 from 2 to 10: 2, 4, 5, 8 and 10. The numbers 3, 6, 7 and 9 are not factors.

a. Start from your birth year: the leap years are the multiples of 4 from then till now (no century year falls between 2001 and 2099).

b. From 2024 till 2099 there are 19 leap years.

Largest = 8888, smallest = 2112. A 4-digit palindrome looks like abba. It is divisible by 4 only if the number formed by its last two digits (ba) is, so the digit a must be even.

a. Sometimes true.

b. Sometimes true.

You do not need long division at all — just look at the last digit.

• By 10: the remainder is the units digit.
• By 5: take the units digit and subtract 5 if it is 5 or more.
• By 2: the remainder is 0 for an even units digit and 1 for an odd one.

The two numbers could be 8 and 5 (or 8 and 10).

Divisible by 8 → also by 4 and 2 (8 = 4 × 2). Divisible by 2 and 5 → also by 10 (10 = 2 × 5).

Being divisible by 2, 4, 5, 8 and 10 all together is the same as being divisible by 40 (= 8 × 5). So check two things: last three digits divisible by 8, and units digit 0 or 5 — in fact 0, since the number must also be even.

Answer: 5600, 6000 and 77622160.

Answer: 16 and 625.

Here is how to find it. Start with the prime factorisation:

Every one of the four can be called special — it depends on what you look at. Here are the reasons Guna’s classmates gave, and a few more.

(one reason for each number — other correct reasons are also fine):

Comparing the two figures gives the rules at once:

Check the solved example given in the book:

Prime factorise every total first, then look for a row or column that is forced.

Puzzle 1 — rows 105, 20, 30; columns 28, 125, 18

Puzzle 2 — rows 8, 105, 70; columns 30, 70, 28

Puzzle 3 — rows 63, 27, 190; columns 45, 42, 171

Puzzle 4 — rows 343, 66, 44; columns 28, 154, 231