Question

5. The smallest number which when divided by 28 and 42 leaves remainders 8 and 22 respectively is: (D) none of these (A) 104 (C) 64 (B) 84 (D) 74​

Answer 1

Answer:The smallest number when divided by 28 and 32 leaves remainder 8 and 12 respectively:-

28 – 8 = 20 and 32 – 12 = 20 are divisible by the required numbers.

Therefore, the required number will be 20 less than the LCM of 28 and 32.

Prime factorization of 28 = 2 x 2 x 7

Prime factorization of 32 = 2 x 2 x 2 x 2 x 2

LCM(28, 32) = 2 x 2 x 2 x 2 x 2 x 7 = 224.

Therefore, the required the smallest number = 224 – 20 = 204.

Verification:

204/28 = 28 x 7 = 196.

= 204 – 196 = 8

204/32= 32 x 6 = 192

= 204 – 192 = 12.

Step-by-step explanation: