Which of the following is the smallest positive integer that when divided by 4, 6 and 8 leaves a remainder of '2' in each case, but is perfectly divisible by 37?
This one is a number properties question and tests your understanding of test of divisibility of numbers and your understanding of the LCM concept.
Let the least positive integer be "x".
"x" leaves a remainder of 2 when divided by 4, 6 and 8.
Therefore, (x - 2) will be divisible by all the three numbers viz., 4, 6 and 8. In other words, (x - 2) is a multiple of 4, 6, and 8 or (x - 2) is a common multiple of 4, 6, and 8.
The least value that x - 2 can take will be the LCM of 4, 6 and 8, which is 24.
The other values that x - 2 can take will be multiples of 24 as all the multiples of 24 will be multiples common to 4, 6 and 8.
Now, if x - 2 = 24, then x = 26. It is not a multiple of 37.
The next value that (x - 2) can be is 48. Then x = 50. 50 is also not a multiple of 37.
The next value that (x - 2) can be is 72. Then x = 74. 74 is a multiple of 37.
Therefore, the correct answer is 74, choice D.
The best approach is to go from answer choices.
You know that the given number should be a multiple of 37 as it is divisible by 37.
The only choice left is choice D.