The question given below is a sample question in Numbers and operation. It is about test of divisibility of numbers. There are usually few questions based on the concept of test of divisibility.
Question
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?
A. 146
B. 666
C. 296
D. 74
E. 242
Correct answer is 74. Choice D.
Explanatory Answer
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.
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.
If x - 2 = 48, then x = 50. It is also not a multiple of 37.
If x - 2 = 72, then x = 74. It is a multiple of 37.
Therefore, the correct answer is 74, choice D.
How to approach this question in the test?
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 - the value suggested for x is 296. This number is divisible by 4 and will therefore, not leave a remainder of "2" when divided by 4. So eliminate it.
The only choice left is choice D.
