Question

Find the smallest number which leaves remainders 8 and 12 when divided by 28 and
32 respectively.
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Answer 1

Answer:

Here we need to find just a single smallest number N ,

such that: when N/28, remainder= 8

& when N/32, remainder= 12

We don't require distinct numbers N1 & N2. But we need to find a single smallest number N

So, First we calculate the smallest number ,which when divided by 28 & 32 leaves remainder 0 in both the cases

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 ●●●●●●●●●●●(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

Justification: 204 divided by 28, remainder = 8

& 204 divided by 32, remainder = 12

Step-by-step explanation: