NCERT Solutions Ganita Prakash Chapter 3 In-text Questions — The Magic Number of Kaprekar

Book page 63 Updated on2026-09-05

Q1.
Take different 4-digit numbers and try carrying out these steps. Find out what happens. Check with your friends what they got. (Also complete the last column: A = ___, B = ___, C = ___)
Answer

Whatever 4-digit number you start with (as long as at least two of its digits are different), you always land on 6174 — the Kaprekar constant — and then it repeats for ever.

The chain started in the book with 6382 finishes like this:

Round1234
A (largest)8632664276417641
B (smallest)2368246614671467
C = A − B6264417661746174

So the last column is A = 7641, B = 1467, C = 7641 − 1467 = 6174. Once you reach 6174 the steps only give 6174 again and again.

Try a fresh number, say 3524:

5432 − 2345 = 3087
8730 − 3078 = 5652
6552 − 2556 = 3996
9963 − 3699 = 6264
6642 − 2466 = 4176
7641 − 1467 = 6174
Try This: ask four friends to each pick a different 4-digit number. Everyone will end at 6174 — only the number of rounds will differ. (A number like 3333, with all four digits the same, gives 0 and is not allowed.)
Q2.
Carry out these same steps with a few 3-digit numbers. What number will start repeating?
Answer

With 3-digit numbers you always reach 495, and then it repeats.

Start with 321:

321 − 123 = 198
981 − 189 = 792
972 − 279 = 693
963 − 369 = 594
954 − 459 = 495
954 − 459 = 495  ← it repeats

Start with 517:

751 − 157 = 594
954 − 459 = 495

Answer: the number 495 starts repeating. It is the Kaprekar constant for 3-digit numbers, just as 6174 is for 4-digit numbers.

Tip: the digits of the starting number must not be all the same (111, 222 … give 0).
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