AP ICET 2017 Shift 1 Paper

Question : 83 of 200

Marks: +1, -0

100≤n≤999

and n leaves remainders 1 and 3 respectively when divided by 3 and 8 then the maximum value of n is

Solution:

LCM (3, 8) = 24
The largest number between 100 and 999 (inclusive) divisible by 24 is 960 = 24k

N/8=3

remainder : 3, 11, 19, 27, ...

N/3=1

remainder : 1, 4, 7, 10, 13, 16, 19 , ...
Required number = 960 + 19 = 979

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