Question

Find the smallest positive integer which when divided by 4, 5, 8, 9 leaves remainder 3, 4, 7, 8 respectively.

Answer 1

Few observations:

1) When number is divided by 4 leaves 3 as remainder therefore
number is 4n+3, n is a integer
2) divided by 5 leaves remainder 4
number is 5m+4, m is a integer .this means number will end with either 9 or 4
3) if number end with 4
then 4n+3=A4(A is digit)
4n=A1 so A1 is odd hence number cannot end with 4
4) Number ends with 9 only
it can be 19,39,59,79....(only satisfies when divided by 4 and 5)
(now next condition)
5) number leaves 7 when divided by 8
number is 8p+7
6) as above conditions number ends with 9 and have odd digit at tens place, let be A9, A is odd digit.
8p+7=A9
8p=A2
number is 7 + ( the number divisible by 8 ending with 2)
i.e. 7+32=39,
7+ 72=79.....(AP with d=40)
7) number leaves 8 as remainder when divisible by 9
number is 9q+8
9q+8=A9
9q=A1
means number is 8+(number divisible by 9 ending with 1)
it can be 81+8=89
179
269....(AP with d=90)
8) point 6 and 7 we have to take a common smallest number
Tn= 39+40n
Tn= 89+90m
39+40n=89+90m
4n=9m+5
find n and m by hit and trial both are natural number.
m=3, n=8
number is 359

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Hope it helped..I tried by best to explain.