Question

find the smallest number which leaves remainder reminder 8 and 12 when divided by 28 and 32 respectively​

Answer 1

Hey mate☺

Given that 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 = 2×2×7

Prime factorization of 32 = 2×2×2×2×2

LCM(28,32) = 2×2×2×2×2×7= 224.

Therefore the required smallest number = 224-20 = 204.

Hope it helps you ☺✌✌

Answer 2

Answer:

204

Step-by-step explanation:

let that number be a

so ATQ

a-8 is divisible by 28

hence( (a-8)+28) that is a+20 must be divisible by 28

and

a-12 is divisible by 32

so((a-12)+32) that is also a+20 must be divisible by 32

from here we get that

a+20 is divisible by both 28 and 32

and as a is the smallest number divisible 28 and 32 leaving remainder

8 and 12 respectively

hence a+20 must be the smallest number divisible by both 28 and 32

that is a+20 is LCM of 28 and 32

now,

28=7*2^2

32=2^5

LCM = 2^5 *7=224

now we get that

a+20=224

so

a=204

hence the smallest number divisible by both 28 and 32 leaving remainder 8 and 12 respectively is 204

I HOPE THIS WILL HELP

THANKYOU