NCERT Solutions for Class 7th Maths Chapter 4 In-text Questions — Does This Ever End? · A Magic Number: 142857

Book page 85 – 86 Updated on2026-09-19

Q1.

Can you find the quotients of 10 ÷ 9, and 100 ÷ 11?

Answer

Both quotients repeat for ever.

10 ÷ 9: 9 goes into 10 once, remainder 1.
10 Tenths ÷ 9 → 1 Tenth, remainder 1 … the same step every time.
10 ÷ 9 = 1.111…

100 ÷ 11: 11 × 9 = 99, remainder 1.
10 Tenths ÷ 11 → 0 Tenths, remainder 10
100 Hundredths ÷ 11 → 9 Hundredths, remainder 1 … now it repeats.
100 ÷ 11 = 9.0909…

Why it happens: in each case a remainder comes back that has already appeared. Once a remainder repeats, every step after it must repeat too, because the same remainder with the same divisor gives the same next digit. 9 and 11 are not built from 2s and 5s, so neither divides any power of ten and neither division can stop.

Q2.

Now divide 1 by 7 (1 ÷ 7). Will this end? Note all the remainders we get. It starts with 1, then 3, then 2, then 6, and so on. What do you observe? Can you explain why this division never ends?

Answer

It never ends. 1 ÷ 7 = 0.142857 142857 14…

Step	Divide	Quotient digit	Remainder
1	10 ÷ 7	1	3
2	30 ÷ 7	4	2
3	20 ÷ 7	2	6
4	60 ÷ 7	8	4
5	40 ÷ 7	5	5
6	50 ÷ 7	7	1 ← back to the start

The remainders run 1 → 3 → 2 → 6 → 4 → 5 → 1 → 3 … — a closed chain of six.

Why it happens: when you divide by 7, the only possible remainders are 0, 1, 2, 3, 4, 5 and 6. A remainder of 0 would end the division, so at most six remainders can occur. With only six choices, a remainder must sooner or later come round again — here after exactly six steps. And the moment a remainder repeats, the whole block of quotient digits repeats with it. That is why not only the remainders but also the digits 142857 cycle for ever.

Tip: the same argument shows that every division of whole numbers either stops or starts repeating — the remainders can never all be new for ever.

Q3.

A Magic Number: 142857. Let us consider the number 142857 that arose when dividing 1 by 7. Multiply 142857 by numbers from 1 to 6. What are the products? What do you notice? Multiply 142857 by 7. What do you observe?

Answer

You get the same six digits back, only rolled around to a different starting point.

Multiplication	Product	What happened
142857 × 1	142857	starts at 1
142857 × 2	285714	starts at 2
142857 × 3	428571	starts at 4
142857 × 4	571428	starts at 5
142857 × 5	714285	starts at 7
142857 × 6	857142	starts at 8
142857 × 7	999999	all nines!

Why it happens: 142857 is the repeating block of 1 ÷ 7. Dividing 2 by 7, 3 by 7 and so on starts the very same chain of remainders at a different place, so the same six digits appear in the same cyclic order. And 7 × 142857 = 999999 because 1/7 + 2/7 + 4/7 … in fact 7 × (1/7) = 1, and 0.142857142857… × 7 = 0.999999… = 1.

Did you know? Numbers like this are called cyclic numbers. 142857 is the smallest one.

Q4.

Try This: To find one such number, you can find 1 ÷ 17 in decimal, and use the repeating block of digits.

Answer

Divide 1 by 17 by long division. The remainders take 16 steps to return to 1.

1 ÷ 17 = 0.0588235294117647 0588235294117647 …

So the repeating block is the 16-digit number 0588235294117647. It is cyclic, exactly like 142857:

588235294117647 × 2 = 1176470588235294
588235294117647 × 3 = 1764705882352941

Each product uses the same digits, started at a different point in the cycle.

Why it happens: dividing by 17 can leave only the remainders 1 to 16. Here all sixteen actually occur before the chain closes, so the repeating block is as long as it possibly can be. Primes that behave like this always produce cyclic numbers.

Tip: 7 and 17 work, but 11 does not — 1 ÷ 11 = 0.0909…, a block of only two digits, far shorter than 10. Try 19 and 23 next; both give full-length cycles.

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