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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

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B

525

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C

535

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D

545

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Solution

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.


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