Question

Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

Answer 1

SOLUTION :  

Given : Two numbers 28 and 32 .

Remainders : 8 & 12.

First, we need to find the LCM of (28 & 32) and then subtract the remainders (8+12=20) from the LCM.

The prime factors of 28 & 32 are :  

28 = 2 x 2 x 7 = 2² × 7

32 = 2 × 2 x 2 × 2 x 2  = 2^5

L.C.M of 28 and 32 = 2^5 x 7 = 224

[LCM of two or more numbers = product of the greatest power of each prime factor involved in the numbers, with highest power.]

Here, 224 is the least number which exactly divides 28 and 32 . But we want the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

Therefore = 224 – (8 +12) = 224 - 24 = 204

Hence,  204 is the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Solutions :-

Find the LCM of 28 and 32 :-

Prime Factors of 28 and 32

28 = 2 × 2 × 7 = 2² × 7
32 = 2 × 2 × 2 × 2 × 2 = 2^5

LCM of 28 and 32 = 2^5 × 7 = 224


Now,
Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

224 - (8 + 12)
= 224 - 20
= 204



Hence,
The smallest number is 204