NCERT Solutions Ganita Prakash Chapter 3 Digit Detectives & Section 3.5 Pretty Palindromic Patterns — In-text Questions

Book page 61 & 62 Updated on2026-09-05

Q1.

Among the numbers 1–100, how many times will the digit ‘7’ occur? Among the numbers 1–1000, how many times will the digit ‘7’ occur?

Answer

From 1 to 100 → the digit 7 occurs 20 times. Count place by place:

In the ones place: 7, 17, 27, 37, 47, 57, 67, 77, 87, 97 → 10 times
In the tens place: 70, 71, 72, 73, 74, 75, 76, 77, 78, 79 → 10 times
Total = 10 + 10 = 20

Note that 77 is counted twice — and correctly so, because Dinesh writes the digit 7 twice while writing 77.

From 1 to 1000 → the digit 7 occurs 300 times.

Think of every number from 000 to 999 written with 3 digits.
That is 1000 numbers × 3 digit-places = 3000 digit-places.
Each of the ten digits 0–9 fills the same share: 3000 ÷ 10 = 300
(1000 itself has no 7, so it changes nothing.)

Another way: in the ones place a 7 appears 100 times, in the tens place 100 times and in the hundreds place 100 times (700–799). Total 100 + 100 + 100 = 300.

Q2.

The numbers 121, 313, 222 are some examples of palindromes using the digits ‘1’, ‘2’, ‘3’. Write all possible 3-digit palindromes using these digits.

Answer

A 3-digit palindrome looks like aba — the first and last digits must be the same. Here a and b can each be 1, 2 or 3.

3 choices for a × 3 choices for b = 9 palindromes

First & last digit	Middle digit 1	Middle digit 2	Middle digit 3
1	111	121	131
2	212	222	232
3	313	323	333

Answer: 111, 121, 131, 212, 222, 232, 313, 323, 333 — nine in all.

Q3.

Try the same procedure for some other numbers, and perform the same steps. Stop if you get a palindrome. Are there numbers for which you do not reach a palindrome at all?

Answer

Try the reverse-and-add steps yourself:

34 + 43 = 77 (1 step)
29 + 92 = 121 (1 step)
48 + 84 = 132, 132 + 231 = 363 (2 steps)
76 + 67 = 143, 143 + 341 = 484 (2 steps)
59 + 95 = 154, 154 + 451 = 605, 605 + 506 = 1111 (3 steps)

For 2-digit starting numbers you always reach a palindrome — some just take longer. The hardest one is 89, which needs 24 steps before it becomes the palindrome 8813200023188!

Did you know? For 3-digit numbers nobody knows the answer. Mathematicians suspect that starting with 196 you never reach a palindrome, even after millions of steps — but this has never been proved.

Q4.

Will reversing and adding numbers repeatedly, starting with a 2-digit number, always give a palindrome? Explore and find out.

Answer

Yes. Every 2-digit number reaches a palindrome after a few reverse-and-add steps.

Start	Steps	Palindrome reached
12	12 + 21	33
47	47 + 74	121
68	68 + 86 = 154; 154 + 451 = 605; 605 + 506	1111
86	86 + 68 = 154; then as above	1111
89	89 + 98 = 187; … 24 steps in all	8813200023188

A short cut: if the two digits of the number add up to 9 or less (like 34 → 3 + 4 = 7), the very first addition already gives a palindrome. That is because there is no carrying: 34 + 43 = 77.

Math Talk: the two numbers 68 and 86 reach the same palindrome, and by the same route. Can you see why? Reversing a number and adding gives the same result whichever of the two you start from.

Q5.

Puzzle time. I am a 5-digit palindrome. I am an odd number. My ‘t’ digit is double of my ‘u’ digit. My ‘h’ digit is double of my ‘t’ digit. Who am I? _________________ Write the number in words:

Answer

Write the number as tth th h t u. Because it is a palindrome, the digits must mirror each other:

tth = u and th = t
So the number looks like u t h t u

Now use the clues one by one:

The number is odd, so the ones digit u is 1, 3, 5, 7 or 9.
t = 2 × u and h = 2 × t = 4 × u. Since h must stay a single digit, 4 × u ≤ 9.
Only u = 1 works: u = 3 would give h = 12, which is not a digit.

u = 1 → t = 2 × 1 = 2 → h = 2 × 2 = 4
Number = 1 2 4 2 1 = 12,421

tth	th	h	t	u
1	2	4	2	1

Answer: 12,421. In words — twelve thousand four hundred twenty-one.

Check it yourself: 12421 read backwards is 12421 ✓ palindrome; it ends in 1 ✓ odd; t = 2 = 2 × 1 ✓; h = 4 = 2 × 2 ✓.

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