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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Let the integers be x - 1, x and x + 1 where x is a positive integer. Since their sum, x - 1 + x + x + 1 = 3x is a multiple of 10, x must be a multiple of 10.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
The median, x, of the three integers is a multiple of 9 and a multiple of 10.
Since x < 100, we must have x = 90.
Therefore, the smallest of the integers is x - 1 = 89.
Thus, condition 1) is sufficient.

Condition 2)
If x = 30, the integers are 29, 30, 31 and the smallest integer is 29.
If x = 90, the integers are 89, 90, 91 and the smallest integer is 89.
Since we do not obtain a unique answer, condition 2) is not sufficient.


Therefore, the answer is A.

Answer: A

Each of three consecutive positive integers is less than 100. Their sum of is a multiple of 10. What is the smallest of the three integers?

1) Their median is a multiple of 9
2) Two of the integers are prime numbers.

We need to determine the value of the smallest of the three integers given that they are consecutive positive integers less than 100 and their sum is a multiple of 10.

If the sum of the three consecutive integers is a multiple of 10, then the middle integer must be a multiple of 10. We can verify this by letting the middle integer be x, and thus the smallest integer will be x - 1 and the largest will be x + 1. We can say that for some positive integer k, we have:

x - 1 + x + x + 1 = 10k

3x = 10k

x = 10k/3

Since x is an integer and 10 is not divisible by 3, then k must be divisible by 3. Thus k/3 is an integer and hence x = 10k/3 = 10(k/3) is a multiple of 10.

Statement One Alone:

Their median is a multiple of 9.

In our stem analysis, we see that the middle integer must be a multiple of 10. However, the middle integer is also the median and we are given that it's a multiple of 9. Thus, the middle integer, i.e., the median, must be both a multiple of 10 and 9. The only positive integer less than 100 that is both a multiple of 10 and 9 is 10 x 9 = 90. Thus, the smallest integers must be 90 - 1 = 89. Statement one alone is sufficient.

Statement Two Alone:

Two of the integers are prime numbers.

Again, using the fact that the middle integer is a multiple of 10, we can say that the possible sets of 3 integers are:

{9, 10, 11}, {19, 20, 21}; {29, 30, 31}, {39, 40, 41}, {49, 50, 51}, {59, 60, 61}, ...

We can stop at {59, 60, 61} since we see that there are two sets that have two of the 3 integers which are prime numbers: {29, 30, 31} and {59, 60, 61}. Thus, the smallest integer could either be 29 or 59. Since we don't have a unique integer, statement two is not sufficient.

Answer: A