M7MBA wrote:If the positive integer n is greater than 6, what is the remainder when n is divided by 6?
(1) When n is divided by 9, the remainder is 2.
(2) When n is divided by 4, the remainder is 1.
Here's another (longer!) approach (MItch's solution is much nicer, but this one will also work)
Target question: What is the remainder when n is divided by 6?
Given: Positive integer n is greater than 6
Statement 1: When n is divided by 9, the remainder is 2.
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When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
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So, from statement 1, we can conclude that some possible values of n are:
11, 20, 29, 38, 47, 56, 65. . . (since we're told n > 6, we can rule out n = 2) .
Let's test some possible values of n:
Case a: n = 11.
When we divide 11 by 6, the remainder is 5
Case b: n = 20.
When we divide 20 by 6, the remainder is 2
Since we cannot answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: When n is divided by 4, the remainder is 1.
from statement 2, we can conclude that some possible values of n are:
9, 13, 17, 21, 25, 29, 33, 37, 41, . . . (since we're told n > 6, we can rule out n = 1 and n = 5) .
Let's test some possible values of n:
Case a: n = 9.
When we divide 9 by 6, the remainder is 3
Case b: n = 13.
When we divide 13 by 6, the remainder is 1
Since we cannot answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 tells us that some possible values of n are:
11, 20, 29, 38, 47, 56, 65, . . .
Statement 2 tells us that some possible values of n are:
9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69 . . .
From the above, we can see that some values of n that satisfy BOTH statements are:
29, 65, etc
We might also notice that if we keep adding 36 (the least common multiple of 9 and 4) to these possible n-values, we'll get more possible n-values such as
101, 137, 173, etc
Let's test some possible values of n:
Case a: n = 29.
When we divide 29 by 6, the remainder is 5
Case b: n = 65.
When we divide 65 by 6, the remainder is 5
Case c: n = 101.
When we divide 101 by 6, the remainder is 5
Case d: n = 137.
When we divide 137 by 6, the remainder is 5
So, it certainly looks like
the remainder will always be 5.
Since we can answer the
target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent
