Question

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

Answer 1

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.


Verification:

204/28 = 28 * 7 = 196.

             = 204 - 196   
 
             = 8


204/32 = 32 * 6 = 192

             = 204 - 192

             = 12.


Hope this helps!

Answer 2

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

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

=> 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

=> 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

●ANS● 204.