1.

Find the least number which when divided by 16, 18, 20 and 25 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.

Answer» First we have to find the LCM of `16,18,20 and 25`.
`16 = 2^4`
`18 = 2*3^2`
`20= 2^2*5`
`25 = 5^2`
so, LCM of these numbers will be `=2^4*3^2*5^2 = 3600`
Let the required number is `x`.
Then, `x = n**3600+4`
`=>x = n**(514**7+2)+4`
`=>x = 514**7n + (2n+4)`
If we divide `x` by `7`, then remainder should be `0`.
`:. (2n+4)/7` should have remainder `0`.
Minimum value for `n` to meet this condition is `5`.
So, required number will be `= 5**3600+4 = 18004`


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