CBSE Class 10 Answered
find the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively
Asked by ajayrath7 | 31 Aug, 2019, 11:38: AM
Expert Answer
The smallest number when divided by 28 and 32 leaves remainder 0 will be the lcm
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.
Prime factorization of 32 = 2 × 2 × 2 × 2 × 2
LCM(28,32) = 2 × 2 × 2 × 2 × 2 × 7
= 224.
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)
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
Answered by Arun | 31 Aug, 2019, 11:59: AM
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