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I chose answer B but I noticed that E can also be an answer to this problem.


42 x 8 = 336, which also leaves a remainder of 1 when divided by 5. Please explain.
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I chose answer B but I noticed that E can also be an answer to this problem.


42 x 8 = 336, which also leaves a remainder of 1 when divided by 5. Please explain.


We have to choose the least possible 3 digit no. That's why.
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Bunuel
What is the least positive three-digit integer that is divisible by both 3 and 14 and leaves a remainder of 1 when divided by 5?

A. 112
B. 126
C. 141
D. 156
E. 336

Divisible by 3 and 14 -> Divisible by the LCM of 3 and 14 -> Divisible by 42.

We want a multiple of 42 that will leave the remainder of 1 when divided by 5, this means the unit digit of 42*i must be 1 or 6. However since 42 is even, we can only aim for the unit digit of 6. The smallest we can go is 42*3 to achieve the unit digit of 6, which is 126. The other options would be 42*8, 42*13 etc.

Ans: B
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Wait, am I missing something? How come is this problem a distance/Rate problem?
Might need to re-label unless I am wrong.
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Bunuel
What is the least positive three-digit integer that is divisible by both 3 and 14 and leaves a remainder of 1 when divided by 5?

A. 112
B. 126
C. 141
D. 156
E. 336
Solution:

The integers that are divisible by both 3 and 14 are the multiples of the LCM of 3 and 14, or 42. Therefore, these integers are 42, 84, 126, and so on. We see that 126 is the first (or least) 3-digit integer that is divisible by both 3 and 14 and also leaves a remainder of 1 when divided by 5, so 126 is the correct answer.

Answer: B
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