In-text — Cryptarithms — Number Play

1. 14 marksGanita Prakash Cl-8 Part 1, in-text, page 131

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

   Hint. Always start at the units column, and track the carry into the next column.

   The rules: each letter is one digit, different letters are different digits, and no number starts with 0. Work from the units column and follow the carry.

   (i) A1 + 1B = B0 Units: 1 + B must end in 0. Since B is a single digit, 1 + B = 10, so B = 9, carrying 1. Tens: A + 1 + 1 (carry) = B = 9, so A = 7. Check: 71 + 19 = 90 ✓

   (ii) AB + 37 = 6A Tens: A + 3 + (carry) = 6. If the carry were 0 then A = 3, and the units would need B + 7 = 3, impossible for a digit. So the carry is 1, giving A + 3 + 1 = 6 and A = 2. Units: B + 7 = 12 (ending in A = 2, with the carry of 1), so B = 5. Check: 25 + 37 = 62 ✓

   (iii) ON + ON + ON = PO, i.e. 3 × ON = PO Write it out: 3(10·O + N) = 10·P + O, so 30O + 3N = 10P + O, giving 29O + 3N = 10P. Since PO is only two digits, 3 × ON < 100, so ON < 34 and O is 1, 2 or 3. Trying O = 3: 87 + 3N = 10P. For the left side to be a multiple of 10, 3N must end in 3, so N = 1, giving 90 = 10P and P = 9. So N = 1, O = 3, P = 9. Check: 31 × 3 = 93 ✓ (and the tens digit of the answer, 9, is P while the units, 3, is O ✓)

   (iv) QR + QR + QR = PRR, i.e. 3 × QR = PRR The answer has three digits, so 3 × QR ≥ 100 and QR ≥ 34. The units of the answer is R, and the units of 3 × QR is the last digit of 3R. So 3R must end in R, which happens only for R = 0 or R = 5. R = 0 would make the answer end in 00, forcing QR to be a multiple of 100 — impossible. So R = 5, and 3 × 5 = 15 carries 1. Now QR = Q5 and 3 × Q5 = P55. Trying Q = 8: 85 × 3 = 255 ✓, giving P = 2. So Q = 8, R = 5, P = 2. Check: 85 + 85 + 85 = 255 ✓

   ✦ (i) A = 7, B = 9. (ii) A = 2, B = 5. (iii) N = 1, O = 3, P = 9. (iv) Q = 8, R = 5, P = 2.

2. 25 marksGanita Prakash Cl-8 Part 1, in-text, page 132

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

   Hint. Look at the units digit of the product first — it usually pins down one letter immediately.

   (i) UT × 3 = PUT The product has three digits while UT has two, so P is small; and the units digit of 3 × T must be T itself, which forces T = 0 or T = 5. If T = 5, then 3 × 5 = 15 carries 1, and the tens of the product would be 3U + 1 ending in U — no single digit works. So T = 0. Then 3 × U must end in U with no carry, giving U = 0 (taken) or U = 5. UT = 50, and 50 × 3 = 150, so P = 1. Check: 50 × 3 = 150 ✓

   (ii) AB × 5 = BC A two-digit number times 5 stays two digits, so AB ≤ 19, meaning A = 1. The product ends in C and starts with B. Since any multiple of 5 ends in 0 or 5, C is 0 or 5. Trying AB = 19: 19 × 5 = 95, so B = 9 ✓ (matching the B in AB) and C = 5. So A = 1, B = 9, C = 5. Check: 19 × 5 = 95 ✓

   (iii) L2N × 2 = 2NP Both sides are three digits, so L2N < 500 and doubling keeps it under 1000. The product starts with 2, so L2N is between 100 and 149, forcing L = 1. So the number is 12N and 2 × 12N = 2NP. Doubling 120 gives 240, doubling 129 gives 258 — so the tens digit N of the product runs from 4 to 5. Trying N = 5: 125 × 2 = 250, so the product is 2NP = 250 with N = 5 ✓ and P = 0. So L = 1, N = 5, P = 0. Check: 125 × 2 = 250 ✓

   (iv) XY × 4 = ZX A two-digit number times 4 stays two digits, so XY ≤ 24, meaning X is 1 or 2. The product ends in X, and 4 × Y is even, so X must be even — hence X = 2. So XY = 2Y and 4 × 2Y ends in 2. Since 4 × Y ends in 2, Y = 3 or 8. But XY ≤ 24 rules out 28, so Y = 3. Then 23 × 4 = 92, giving Z = 9. Check: 23 × 4 = 92 ✓

   (v) PP × QQ = PRP PP and QQ are repeated-digit numbers, so PP = 11P and QQ = 11Q. The product is a three-digit number, which keeps things small. Since the product PRP starts and ends with P, and the units digit of the product is the last digit of P × Q, we need P × Q to end in P. Trying Q = 1 makes QQ = 11, and PP × 11 = 11P × 11 = 121P. For P = 2: 22 × 11 = 242, which fits the pattern PRP = 242 with P = 2 and R = 4 ✓ So P = 2, Q = 1, R = 4. Check: 22 × 11 = 242 ✓

   (vi) JK × 6 = KKK KKK is a three-digit repdigit, so KKK = 111 × K. Since 111 = 3 × 37, we need JK × 6 = 111K, giving JK = 111K ÷ 6 = 18.5K. For JK to be a whole number, K must be even. Testing: K = 4 gives JK = 74 ✓ (a two-digit number ending in 4 ✓). So J = 7, K = 4. Check: 74 × 6 = 444 ✓

   ✦ (i) U = 5, T = 0, P = 1. (ii) A = 1, B = 9, C = 5. (iii) L = 1, N = 5, P = 0. (iv) X = 2, Y = 3, Z = 9. (v) P = 2, Q = 1, R = 4. (vi) J = 7, K = 4.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp105.pdf). The chapter is about divisibility reasoning, digital roots and cryptarithms — questions sit in four 'Figure it Out' blocks (pages 122, 126, 131 and 132-135) plus several in-text drills. Every numeric answer here was independently recomputed before being compared with the book's printed answer key: the full 10-row divisibility table was re-derived rule by rule, every cryptarithm was re-solved from scratch, and the divisibility-by-44 and multiple-of-18 digit pairs were found by exhaustive search over the digits.. Questions are referenced from the NCERT textbook for identification.