Bunuel
n is an integer greater than 1. If the remainder is 1, when n is divided by 3, then (n² + n - 2) must be divisible by which of he following double-digit number?
A. 12
B. 14
C. 16
D. 18
E. 20
There's a nice rule that says, "
If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3
The question tells us that,
when n is divided by 3, the reminder is 1. Notice that the question doesn't tell us what the quotient is.
So, let's assign a variable to the quotient.
That is, we can rephrase the question as:
when n is divided by 3, the quotient is k, and the reminder is 1. We can now apply the above
property to write: n = 3k + 1
The question: (n² + n - 2) must be divisible by which of he following double-digit number? Since n = 3k + 1, we can substitute 3k + 1 for n to get: n² + n - 2 = (3k + 1)² + (3k + 1) - 2
= 9k² + 6k + 1 + 3k + 1 - 2
= 9k² + 9k
At this point we need to determine the biggest integer that must divide into 9k² + 9k
First recognize that we can factor this to get: 9k² + 9k = 9(k² + k)
So we can definitely see that 9 is a divisor of 9(k² + k), but this isn't among the answer choices.
Now recognize that we can also factor k² + k to get: 9(k² + k) = 9(k)(k + 1)
Since k and k + 1 are CONSECUTIVE integers, we know that one of the values must be odd, and the other value must be even.
Since one of those values (k or k+1) must be EVEN we can factor a 2 from one of them.
In other words: 9(k)(k + 1) = 9(2)(something) = (
18)(something)
The expression is clearly divisible by
18Answer: D
Tricky question!!
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