Question

The product of two digit numbers is 1938. If the product of their unit's digit is 28 and that of their ten's digits is 15, find the numbers

Answer 1

Given data :

• product of two digit number = 1938
• product if units digits = 28
• product of tens digits = 15

Solution :

Let, two digit number be (10a + b) and (10c + d)

were, b and c are unit digit and a and c are ten's digit

Now, according to given;

⟹ (10a + b) (10c + d) = 1938

⟹ 100ac + 10ad + 10cd + bd = 1938

Now, from given

• bd = 28
• ac = 15

⟹ 100 * 15 + 10ad + 10cb + 28 = 1938

⟹ 1500 + 10ad + 10cd + 28 = 1938

⟹ 10ad + 10cb = 1938 - 1500 - 28

⟹ 10ad + 10cd = 410 { 1 }

case 1 :

let, b = 4 and d = 7

a = 5 and c = 3

Now put values in eq. { 1 }

⟹ 10 * 5 * 7 + 10 * 3 * 4 = 410

⟹ 350 + 120 = 410

⟹ 470 ≠ 410

case 2:

let, b = 7 and d = 4

a = 5 and c = 3

Now put values in eq. { 1 }

⟹ 10 * 5 * 4 + 10 * 3 * 7 = 410

⟹ 10 * 40 + 10 * 21 = 410

⟹ 400 + 210 = 410

Now put values in eq. { 1 }

⟹ 10 * 5 * 7 + 10 * 3 * 4 = 410

⟹ 350 + 120 = 410

⟹ 410 = 410

Now, put values b = 7 and d = 4, a = 5 and c = 3 in

(10a + b) and (10c + d)

⟹ (10a + b) = 10 * 5 + 7 = 50 + 7 = 57

⟹ (10c + d) = 10 * 3 + 4 = 30 + 4 = 34

Answer : Hence the numbers are 57 and 34.