t^2+5t+6 is begging to be factored, so let's start there:
t^2+5t+6=(t+2)(t+3)
So the question is, "What is the remainder when (t+2)(t+3) is divided by 7?"
Statement 1:If t divided by 7 has a remainder of 6, this means that t is 6 more than a multiple of 7. So, a number like 13 or 20. If this is the case, t+1 is exactly a multiple of 7, t+2 is 1 more than a multiple of 7, and t+3 is 2 more than a multiple of 7. So t+2 divided by 7 has a remainder of 1 and t+3 divided by 7 has a remainder of 2.
In general, if you multiply two numbers together with remainders a and b, the product will have the same remainder as ab does. So in this case, the remainder of (t+2)(t+3) is just 2. SUFFICIENT. One more example to clarify this rule.
When dividing by 7:
12 has a remainder of 5
9 has a remainder of 2
12*9=108 has a remainder of 3
5*2=10 has a remainder of 3
This works when dividing by all integers by the way, not just 7.
Statement 2:
t^2 divided by 7 has a remainder of 1. This can happen if t has a remainder of 6 or if t has a remainder of 1 when divided by 7. For example 6^2=36(remainder 1). 8^2=64(remainder 1). You can also think of a remainder of 6 when dividing by 7 as the same as a remainder of -1. That is, instead of thinking of it as SIX MORE THAN a multiple of 7, think of it as ONE LESS THAN a multiple of 7. For example, 27 is 6 more than a multiple of 7(21) and 1 less than a multiple of 7(28). Thinking of it this way you could apply the above rule easily and reason as follows:
If t^2 has a remainder of 1, then t must have a remainder of -1 or 1, because those are the only numbers you could square to get 1. If t has a remainder of -1, t+2 has a remainder of 1 and t+3 has a remainder of 2, meaning (t+2)(t+3) has a remainder of 2. If t has a remainder of 1, then t+2 has a remainder of 3 and t+3 has a remainder of 4, and (t+2)(t+3) has a remainder of 5. (because 4*3=12 which has a remainder of 5 when divided by 7. INSUFFICIENT
Number Problem in DS
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