Abhi077 wrote:

What is the sum of the two digit numbers that leave a remainder of 1 when divided by both 3 & 4

A)430

B)440

C)450

D)460

E)470

When the number is divided by 3 or by 4, the remainder is 1.

This means the number is of the form

N = 12a + 1

Since 12 is the LCM of 3 and 4.

All such 2 digit numbers are 12*1 + 1, 12*2 + 1, .... 12*8+1

This is an arithmetic progression of 8 terms with first term as 13 and last term as 97.

Sum = n*(First term + Last term)/2 = 8*(13 + 97)/2 = 440

Answer (B)

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Karishma

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