Question

Four numbers leave remainders of 243, 289, 293 and 305 on divisible by T. The sum of the four numbers leaves a remainder of 60 on divisible by T. The value of T can be

• A. 515
• B. 525
• C. 535
• D. 545

Answer 1

The correct option is C

535

The sum of the four numbers should leave a remainder of 1130(243+289+293+305). It should be in the form of Ta + 1130 (Where a is an integer).

However, it leaves a remainder of 60. It should be in the form of Tb + 60.

Ta + 1130 = Tb + 60

T(b - a) = 1070

T should be a factor of 1070.

T should be greater than 305(since 243, 289, 293 and 305 are remainders).

Factors of 1070 that are greater than 305 is 535.

So, the value of T is 535.