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## SAT Numbers Practice Question: Test of Divisibility

### Question 2: Divisibility & LCM

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?

1. 146
2. 666
3. 296
4. 74
5. 242
Choice D. The number is 74.

This one is a number properties question and tests your understanding of test of divisibility of numbers and your understanding of the LCM concept.

### Find the LCM of 4, 6, and 8 and then proceed

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.

### The rest is iterative

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.

• Look at choice A. 146 is not a multiple of 37. Eliminate it. So, is the case for 242. So, you have eliminated choices A and E.
• Now look at choice B. It is evident that 666 is divisible by 6. Therefore, it will not leave a remainder of 2. So, eliminate choice B.
• Choice C - 296. This number is divisible by 4 and will therefore, not leave a remainder of "2" when divided by 4. So eliminate choice C.

The only choice left is choice D.