Question

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

Answer 1

Answer:

Step-by-step explanation:

First we calculate the smallest number ,which when divided by 28 & 32 leaves remainder 0.

Which is the LCM of 28, & 32 =

28= 2^2 * 7

2 = 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

=> 224–20 = 28*7 +8 ----------(1)  

lly, 224 = 32 *7 +0

=> 224 = 224 +0

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

=> 224 = (32*6+20) +12

=> 224 - 20 = 32*6 +12 ------------(2)

Now, we observe that LHS of both eq (1) & eq(2) are equal. Divisors are also 28 & 32 & remainders are also the required numbers 8 & 12

So the smallest dividend = 224 - 20 = 204

--------------dark lover----

Answer 2

Answer:

224

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.

Hope it helps u