NCERT Solutions Ganita Prakash (Part 1) Chapter 5 .3 Digits in Disguise — In-text Questions

Book page 1325 Updated on2026-09-05

Q1.

(vi) Try this now: GH × H = 9K. This means a 2-digit number multiplied by a 1-digit number gives another 2-digit number in the 90s. Observe the letters corresponding to the units digits in this cryptarithm. Pick the solution to this question from the options given below: 11 × 9 = 99, 12 × 8 = 96, 46 × 2 = 92, 24 × 4 = 96, 47 × 2 = 94, 31 × 3 = 93, 16 × 6 = 96.

Answer

The solution is 24 × 4 = 96, giving G = 2, H = 4, K = 6.

The letter to watch is H: it is both the units digit of the two-digit number and the multiplier. So the multiplier must equal the units digit.

Option	Units digit vs multiplier	Fits GH × H = 9K?
11 × 9 = 99	1 ≠ 9	No
12 × 8 = 96	2 ≠ 8	No
46 × 2 = 92	6 ≠ 2	No
24 × 4 = 96	4 = 4 ✓	Yes
47 × 2 = 94	7 ≠ 2	No
31 × 3 = 93	1 ≠ 3	No
16 × 6 = 96	6 = 6 ✓, but then H = 6 and K = 6	No — two letters cannot share a digit

Tip: 16 × 6 = 96 is the near miss worth studying. It passes the units-digit test but breaks the basic rule of a cryptarithm — different letters must stand for different digits.

Q2.

(vii) Here is one more: BYE × 6 = RAY. Anshu says, “Since the product is a 3-digit number, B can’t be 2 or more. If B = 2, i.e., 2 hundreds, the product will be more than 1200. So, B = 1.” What can you say about ‘Y’? What digits are possible/not possible?

Answer

Y must be even and less than 7, so the only possibilities are Y = 0, 2, 4 or 6. Testing them leaves exactly one solution.

Two facts pin Y down straight away:

Y is even. The product RAY ends in Y, and 6 × (anything) is even. So Y cannot be 1, 3, 5, 7 or 9.
Y is less than 7. If Y = 7, then BYE is at least 170, and 170 × 6 = 1020 — a 4-digit product. So 7, 8 and 9 are out.

Now try each surviving value, remembering B = 1 and that 6 × E must end in Y:

Y	E forced by “6 × E ends in Y”	Outcome
0	E = 5 (6 × 5 = 30)	105 × 6 = 630 → R = 6, A = 3 ✓ all letters different
2	E = 7 (6 × 7 = 42)	127 × 6 = 762 → R = 7 clashes with E = 7 ✗
4	E = 9 (6 × 9 = 54)	149 × 6 = 894 → A = 9 clashes with E = 9 ✗
6	E = 1 or 6	both clash — E = 1 is B, E = 6 is Y ✗

So B = 1, Y = 0, E = 5, R = 6, A = 3 and 105 × 6 = 630 ✓

Tip: Y = 0 is allowed because Y is never the leading digit of either number. Only B and R are leading digits, and neither of them may be 0.

Q3.

Solve the following: (i) UT × 3 = PUT (ii) AB × 5 = BC (iii) L2N × 2 = 2NP (iv) XY × 4 = ZX (v) PP × QQ = PRP (vi) JK × 6 = KKK

Answer

(i) U = 5, T = 0, P = 1 → 50 × 3 = 150

3(10U + T) = 100P + 10U + T
30U + 3T = 100P + 10U + T
20U + 2T = 100P → 10U + T = 50P
So UT = 50P; P = 1 gives UT = 50 (P = 2 would need UT = 100)

(ii) A = 1, B = 9, C = 5 → 19 × 5 = 95

A 2-digit × 5 stays 2-digit only if AB ≤ 19, so A = 1
(10 + B) × 5 = 10B + C
50 + 5B = 10B + C → C = 50 – 5B
C must be a digit → 50 – 5B ≤ 9 → B = 9, C = 5

(iii) Two solutions: L = 1, N = 5, P = 0 and L = 1, N = 4, P = 8.

2(100L + 20 + N) = 200 + 10N + P
200L + 40 + 2N = 200 + 10N + P
200L + 40 = 200 + 8N + P
L = 1 → 8N + P = 40 (L = 2 would need 8N + P = 240)

N = 5 gives P = 0 → 125 × 2 = 250 ✓ N = 4 gives P = 8 → 124 × 2 = 248 ✓. Both keep every letter on its own digit and away from the printed 2. (The book's answer key lists only 125 × 2 = 250.)

(iv) X = 2, Y = 3, Z = 9 → 23 × 4 = 92

4(10X + Y) = 10Z + X → 39X + 4Y = 10Z
2-digit product → XY ≤ 24, so X = 1 or 2
X = 1: 39 + 4Y = 10Z needs 4Y to end in 1 — impossible, 4Y is even
X = 2: 78 + 4Y = 10Z needs 4Y to end in 2 → Y = 3 → Z = 9

(v) Three solutions, one of which is P = 2, Q = 1, R = 4.

PP = 11P and QQ = 11Q, so PP × QQ = 121 × P × Q
The product is 3-digit → P × Q ≤ 8
121 × 2 = 242, 121 × 3 = 363, 121 × 4 = 484 — each of the form PRP

Product	Reading it as PRP	Cryptarithm
242	P = 2, R = 4, so Q = 1	22 × 11 = 242 ✓
363	P = 3, R = 6, so Q = 1	33 × 11 = 363 ✓
484	P = 4, R = 8, so Q = 1	44 × 11 = 484 ✓

121 × 1 = 121 fails because it forces P = Q = 1, and 121 × 5 = 605 onwards no longer starts and ends with the same digit.

(vi) J = 7, K = 4 → 74 × 6 = 444

KKK = 111K, so 6(10J + K) = 111K
60J + 6K = 111K
60J = 105K → 4J = 7K
K must be a multiple of 4 → K = 4, J = 7 (K = 8 would give J = 14)

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