NCERT Solutions Ganita Prakash (Part 1) Chapter 6 In-text Questions — Digits in Disguise

Book page 143 Updated on2026-09-05

Q1.

What could U and T be? Can T be 2? Can it be 3?

Answer

The puzzle is T + T + T = UT, that is, 3 × T must be a 2-digit number whose units digit is again T.

T	3 × T	Units digit	Same as T?
2	6	6	No (and 6 is not 2-digit)
3	9	9	No
4	12	2	No
5	15	5	Yes
6	18	8	No
7	21	1	No
8	24	4	No
9	27	7	No

So T = 5 and U = 1, giving UT = 15.

5 + 5 + 5 = 15 ✓

T cannot be 2 (3 × 2 = 6 is not even a 2-digit number) and T cannot be 3 (3 × 3 = 9, again a single digit, and the units digit is 9, not 3).

Why it happens: 3T ending in T means 3T – T = 2T ends in 0, so 2T must be a multiple of 10. For a single digit T that forces T = 0 or T = 5, and T = 0 would not give a 2-digit sum.

Q2.

Here K2 means that the number is a 2-digit number having the digit ‘2’ in the units place and ‘K’ in the tens place. K2 is added to itself to give a 3-digit sum HMM. What digit should the letter M correspond to?

Answer

M = 4, and the full solution is K = 7, H = 1, so 72 + 72 = 144.

Units column: 2 + 2 = 4, so M = 4 with no carry.
Tens column must also show M = 4:
K + K = 4 or 14 → 2K = 14 → K = 7 (2K = 4 gives K = 2, but then 22 + 22 = 44 is not a 3-digit number)
Carry 1 into the hundreds column → H = 1
So 72 + 72 = 144 ✓

Why it happens: the units digit is decided first and it fixes M straight away. The tens digit of the answer must then match that same M, which pins down K.

Q3.

What about H? Can it be 2? Can it be 3?

Answer

H must be 1. It cannot be 2 or 3.

K2 is at most 92, so K2 + K2 is at most 92 + 92 = 184.
A 3-digit sum below 200 must begin with 1.
So H = 1, and H = 2 or H = 3 is impossible.

Why it happens: doubling a 2-digit number can never reach 200. The hundreds digit of the answer can only come from a carry of 1, so H = 1 always.

Q4.

Find out what each letter stands for: YY + Z = ZOO; B5 + 3D = ED5; KP + KP = PRR; C1 + C = 1FF

Answer

Puzzle 1: YY + Z = ZOO

YY is at most 99 and Z at most 9, so the sum is at most 108.
A 3-digit sum at most 108 must be 100 – 108, so Z = 1 and O = 0, giving ZOO = 100.
YY + 1 = 100 → YY = 99 → Y = 9
Check: 99 + 1 = 100 ✓

Puzzle 2: B5 + 3D = ED5

Units: 5 + D must end in 5 → D = 0, no carry.
Tens digit of the answer is D = 0, so B + 3 must end in 0 → B + 3 = 10 → B = 7, carry 1.
Hundreds: E = 1
Check: 75 + 30 = 105 ✓

Puzzle 3: KP + KP = PRR

Doubling a 2-digit number is at most 198, so P = 1.
2 × (10K + 1) = 100 + 11R
20K + 2 = 100 + 11R → 20K – 98 = 11R
K = 6 gives 120 – 98 = 22 = 11 × 2 → R = 2
Check: 61 + 61 = 122 ✓

Puzzle 4: C1 + C = 1FF

(10C + 1) + C = 11C + 1, and 1FF = 100 + 11F
11C + 1 = 100 + 11F → 11(C – F) = 99 → C – F = 9
So C = 9 and F = 0
Check: 91 + 9 = 100 ✓

Cryptarithm	Letters	Sum in figures
YY + Z = ZOO	Y = 9, Z = 1, O = 0	99 + 1 = 100
B5 + 3D = ED5	B = 7, D = 0, E = 1	75 + 30 = 105
KP + KP = PRR	K = 6, P = 1, R = 2	61 + 61 = 122
C1 + C = 1FF	C = 9, F = 0	91 + 9 = 100

Why it happens: always start where the information is tightest — usually the units column (it has no incoming carry) or the leading digit of a longer answer (it can only be a carry, so it is almost always 1).

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