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What is the least positive integer which leaves a remainder of 1 when divided by 3; 2 when divided by 4; and 3 when divided by 5?

We can test the answer choices

Remainder of 3 when divided by 5

=> If number leaves 3 as remainder when divided by 5
=> Units digit should be 3 or 8 (As 3 = 0 + 3 and 8 = 5 + 3)
=> We can ELIMINATE C and E

Remainder of 1 when divided by 3

=> Subtract 1 from the numbers and add the digits of the number to see if we get a multiple of 3

A. 118 - 1 = 117 = 1 + 1 + 7 = 9 => Multiple of 3 => KEEP
B. 68 - 1 = 67 => Not a Multiple of 3 => ELIMINATE
D. 58 - 1 = 57 => Multiple of 3 => KEEP

Remainder of 2 when divided by 4

=> Subtract 2 from the numbers and check if we get a multiple of 4

A. 118 + 2 = 120 => Multiple of 3 => KEEP
D. 58 + 2 = 60 => Multiple of 3 => KEEP

We are looking for the least integer and 58 < 118

So, Answer will be D
Hope it helps!

Watch the following video to MASTER Remainders

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