What will be the East number which when divided by 16 18 20 and 2 leaves 4 as remainder in each case but when divided by 7 leaves no remainder?

Text Solution

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

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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.

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