NCERT Solutions Ganita Prakash (Part 1) Chapter 5 .3 Digits in Disguise

Q1.

Solve the cryptarithms given below. (i) A1 + 1B = B0 (ii) AB + 37 = 6A (iii) ON + ON + ON = PO (iv) QR + QR + QR = PRR

Answer

(i) A = 7, B = 9 → 71 + 19 = 90

Units: 1 + B must end in 0 → B = 9, carry 1
Tens: A + 1 + 1 = B = 9 → A = 7
Check: 71 + 19 = 90 ✓

(ii) A = 2, B = 5 → 25 + 37 = 62

Tens: A + 3 + (carry) = 6
If the carry is 0: A = 3, and then units 3 = B + 7 gives B = – 4 ✗
So the carry is 1: A + 4 = 6 → A = 2
Units: B + 7 = A + 10 = 12 → B = 5
Check: 25 + 37 = 62 ✓

(iii) Three solutions, one of which is O = 3, N = 1, P = 9. The puzzle is 3 × ON = PO with a two-digit product.

Units: 3N must end in O
The product must stay below 100, so ON ≤ 33 → O = 1, 2 or 3

O	N (so that 3N ends in O)	Check
1	7 (3 × 7 = 21)	17 × 3 = 51 = PO ✓ → P = 5
2	4 (3 × 4 = 12)	24 × 3 = 72 = PO ✓ → P = 7
3	1 (3 × 1 = 3)	31 × 3 = 93 = PO ✓ → P = 9

All three keep the letters on different digits, so all three are valid: 17 + 17 + 17 = 51, 24 + 24 + 24 = 72 and 31 + 31 + 31 = 93. (The book's answer key lists only the last one.)

(iv) Q = 8, R = 5, P = 2 → 85 + 85 + 85 = 255

3 × QR = PRR, a three-digit number, so QR ≥ 34
Units: 3R must end in R → 2R ends in 0 → R = 0 or 5
R = 0 forces 30Q = 100P, impossible for digits
So R = 5: 3(10Q + 5) = 100P + 55
30Q + 15 = 100P + 55 → 3Q = 10P + 4
P = 2 gives 3Q = 24 → Q = 8
Check: 85 × 3 = 255 ✓

Q2.

(v) PQ × 8 = RS. Guna says, “Oh, this means a 2-digit number multiplied by 8 should give another 2-digit number. I know that 10 × 8 = 80. But the units digits of 10 and 80 are the same, which we don’t want. For the same reason PQ cannot be 11 as P and Q correspond to different digits. 12 × 8 = 96 fits all the conditions”. Can PQ be 13? Think.

Answer

No, PQ cannot be 13 — and 12 is the only value that works.

13 × 8 = 104, a 3-digit number ✗

For the product to stay at two digits we need PQ × 8 ≤ 99, that is PQ ≤ 12. So only 10, 11 and 12 are candidates:

PQ	PQ × 8	Allowed?
10	80	No — Q and S would both be 0
11	88	No — P and Q would both be 1
12	96	Yes — P = 1, Q = 2, R = 9, S = 6

Why nothing bigger works: multiplication by 8 is increasing, so once 13 × 8 crosses 100, every larger two-digit number does too. One check at the boundary rules out all 87 remaining possibilities at once — that is the economy Guna is using.

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