If s and t are positive integers, st + s + t cannot be

take cases for(s,t)

(1,2)-5
(1,3)-7
(2,2)-8
(1,4)-9

only left is B-6

Thanks for the reply! One question - How did you Pick the number? Was it trial and error, some logic or practice?

Thanks in advance!

The problem is misleading. Of course, if s and t can be the same number, then the answer is 6. But if t and s are different numbers... common, why would they are called s and t (is not that because they are different?) Misleading 100%.

Which "same" number gives the answer 6 ?

The two numbers are different. But hypothetically speaking, even if the two numbers were the same, it's perfectly fine to have two different variables represent the same number.

Solution:

SInce the given choices are “small,” the values of s and t have to be even smaller. So if we let s = 1 and t = 2, we have:

(1)(2) + 1 + 2 = 5

If we keep s = 1 and let t = 3, we have:

(1)(3) + 1 + 3 = 7

If we keep s = 1 and let t = 4, we have:

(1)(5) + 1 + 5 = 9

If we keep t = 2 and let s = 2, we have:

(2)(2) + 2 + 2 = 8

Since 5, 7, 8 and 9 can be written in the form of st + s + t for some positive integers s and t, 6 is the one that cannot be written in such a format.

Alternate Solution:

Let’s rewrite st + s + t as follows:

st + s + t + 1 - 1

s(t + 1) + (t + 1) - 1

(s + 1)(t + 1) - 1

Now, if (s + 1)(t + 1) - 1 is equal to 5, 7, 8, or 9; then (s + 1)(t + 1) is equal to 6, 8, 9, or 10, respectively. In each of these cases, we can find values for s + 1 and t + 1 that are greater than 1; therefore, we can find positive integer values for s and t. However, if (s + 1)(t + 1) - 1 = 6; then (s + 1)(t + 1) = 7. Since 7 is prime, either s + 1 = 1 or t + 1 = 1, which is equivalent to saying either s = 0 or t = 0. Thus, there are no positive integer values for s and t which satisfy (s + 1)(t + 1) - 1 = 6.

Answer: B

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