Question

write the smallest number divisible by 32,64,128 and leaves reminder 3 in each case.​

Answer 1

Answer:

remainder 8 and 12 respectively.

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.

Answer 2

Step-by-step explanation:

Which is the LCM of 28, & 32 =

28= 2² x 7

32 = 2^5

So, LCM = 2^5 * 7 = 224

Now, by Euclid's division lemma ,

dividend = divisor*quotient + remainder (r<divisor)

224 = 28 *8 +0

=> 224 = 224 +0

=> 224 = 216 + 8 ( because we want r= 8)

=> 224 = (28*7+20)+8 ( since divisor given is 28)

=> 224–20 = 28*7 +8 ●●●●●●●●●●(1)

Similarly, 224 = 32 *7 +0

=> 224 = 224 +0

=> 224 = 212 +12 ( because we want r= 12)

=> 224 = (32*6+20) +12 ( since divisor given is 32)

=> 224 - 20 = 32*6 +12 ●