Question

The product on 9. Find the smallest number which when divided by 12, 18 and 36 leaves 5 as remainder in each case. 2. n Find the smallest number which when increased by 3 is exactly divisible by 24 36
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Answer 1

Answer:

LCM of 12,18,36 is 36 . add 5 in 36 =41

41÷12=. q =3 and r=5

41 ÷18=. q =2 and r=5

41÷36 =. q=1and r= 5

So , the smallest number is 41

LCM of 24 and 36 is 72 , subtract 3 from 72

72-3= 69

so the smallest number when increased by 3 is exactly divisible by 24 and 36 is 69.