Question 20
When a heap of pebbles is grouped in 32, 40 or 72 it is left with remainders of 10,
18 or 50 respectively. What is the minimum number of pebbles that the heap contains?
Solution
Family of numbers when divided by a,b,c and leaves remainders x,y,z respectively such that (a-x)=(b-y)=(c-z)=v are of the form N = k *LCM ( a,b,c) - v
Therefore N = k*LCM(32,40,72) - 22 => N=1440k-22
Smallest such number is obtained when k=1 which is 1418
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