Question

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Answer 1

Answer:

MATHS

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

November 22, 2019Anshidha Basak

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ANSWER

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

Which is the LCM of 28 & 32

28=22×7

32=25

So, LCM=25×7=224

Now, by Euclid's division lemma

Dividend=divisor×quotient+remainder (r

224=28×8+0

⇒224=224+0

⇒224=216+8(because we want r=8)

⇒224=(28×8+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. Divisor are also 28&32 ,remainder are also the required nuber 8&12

So the smallest dividend=224−20=204