Is abc/d an integer if a, b, c, and d are positive integers?
(1) (a + b + c)/d is an integer.
(2) {a, b, c, d} are consecutive integers and arranged in ascending order
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Is abc/d an integer if a, b, c, and d are positive integers?
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gmat_winter
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Brent@GMATPrepNow
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Target question: Is abc/d an integer?gmat_winter wrote:Is abc/d an integer if a, b, c, and d are positive integers?
(1) (a + b + c)/d is an integer.
(2) {a, b, c, d} are consecutive integers and arranged in ascending order
Statement 1: (a + b + c)/d is an integer
This statement doesn't FEEL sufficient, so I'm going to TEST some values.
There are several values of a, b, c, and d that satisfy statement 1. Here are two:
Case a: a = 1, b = 1, c = 1 and d = 1. This meets the condition that (a + b + c)/d is an integer. In this case, abc/d = (1)(1)(1)/1 = 1, which IS an integer
Case b: a = 1, b = 1, c = 1 and d = 3. This meets the condition that (a + b + c)/d is an integer. However, in this case, abc/d = (1)(1)(1)/3 = 1/3, which is NOT an integer
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, you can read my article: https://www.gmatprepnow.com/articles/dat ... lug-values
Statement 2: {a, b, c, d} are consecutive integers and arranged in ascending order
This statement doesn't feel sufficient either, so let's TEST some values.
There are several values of a, b, c, and d that satisfy statement 2. Here are two:
Case a: a = 1, b = 2, c = 3 and d = 4. This meets the statement 2 condition. In this case, abc/d = (1)(2)(3)/4 = 6/4 = 3/2, which is NOT an integer
Case b: a = 3, b = 4, c = 5 and d = 6. This meets the statement 2 condition. In this case, abc/d = (3)(4)(5)/6 = 60/5 = 12, which IS an integer
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
If a, b, c, d are consecutive integers and arranged in ascending order, then we can rewrite all 4 integers in terms of a.
That is:
a = a
b = a + 1
c = a + 2
d = a + 3
Statement 1 tells us that (a + b + c)/d is an integer.
This means that [a + a+1 + a+2]/(a+3) is an integer
Simplify to see that (3a + 3)/(a + 3) is an integer
IMPORTANT: I know that (3a + 9) is divisible by (a + 3), so let's take the above expression an rewrite it to get:
(3a + 9 - 6)/(a + 3) is an integer [notice that (3a + 9 - 6) is the same as (3a +3)]
Now rewrite the above to get: (3a + 9)/(a + 3) - 6/(a + 3) is an integer
Simplify to get: 3 - 6/(a + 3) is an integer
From this, we can conclude that 6/(a + 3) is an integer
If n is a POSITIVE integer, what can we conclude about a?
In order for 6/(a + 3) to be an integer, a MUST EQUAL 3
If a MUST EQUAL 3, then we know that the 4 values are: a = 3, b = 4, c = 5 and d = 6
Now that we know all of the values of a, b, c and d, we can answer the target question with certainty.
Answer = C
Cheers,
Brent
Last edited by Brent@GMATPrepNow on Mon Apr 20, 2015 6:01 am, edited 1 time in total.
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MartyMurray
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Brent, in Statement 2, case b, I believe you put a 3 where you mean 4.
On another note, I wonder if one would see a question like this on the actual test.
For one thing, proving Statement 2 insufficient requires finding that one case that creates an integer.
Also, the way Brent did it seems to be the only decent way to go about it. I usually find GMAT quant questions to be more hackable than this one is.
In any case, cool question and cool explanation, definitely something one can learn from.
Maybe the biggest takeaway is the usefulness of translating consecutive integers into n, n + 1, n + 2,...
On another note, I wonder if one would see a question like this on the actual test.
For one thing, proving Statement 2 insufficient requires finding that one case that creates an integer.
Also, the way Brent did it seems to be the only decent way to go about it. I usually find GMAT quant questions to be more hackable than this one is.
In any case, cool question and cool explanation, definitely something one can learn from.
Maybe the biggest takeaway is the usefulness of translating consecutive integers into n, n + 1, n + 2,...
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
Perfect Scoring Tutor With Over a Decade of Experience
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Contact me at [email protected] for a free consultation.
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Matt@VeritasPrep
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I like Brent's approach, but let me supplement it a bit.
S1::
Under test conditions, I'd try numbers. It's easy to find a case in which (a + b + c)/d = integer implies (abc)/d = integer ... just make d = 1.
Now we need an easy case in which (a + b + c) / d = integer doesn't imply abc/d = integer. I'd go for d = an odd prime. (For instance, a = 1, b = 4, c = 7, d = 3.)
S2::
Let's say a = n - 3, b = n - 2, c = n - 1, and d = n. Then we have
(n-3)(n-2)(n-1) / n
We can see that every term in the numerator will have an n in it, except the (-3)*(-2)*(-1), or -6. So the numerator will divide by n if and only if -6 divides by n. So the question becomes "is n a factor of 6?"
Further, we might notice that a, b, c, and d are all positive, and S2 implies that a < b < c < d, so d is at least 4. That means the ONLY possible value of d that's a factor of 6 is d = 6. So the question becomes "Is d = 6?" We can't answer yet, so this isn't sufficient.
S1 + S2::
Using our trick from S2 in conjunction with S1, we get (n - 3) + (n - 2) + (n - 1) is divisible by n, or (3n - 6) is divisible by n. Since n divides the first term (3n) it must also divide the second term (-6), so n is a factor of 6. We also know (from S2) that n is at least 4. Thus n MUST be 6. Hence a = 3, b = 4, c = 5, and d = 6, and we're done.
S1::
Under test conditions, I'd try numbers. It's easy to find a case in which (a + b + c)/d = integer implies (abc)/d = integer ... just make d = 1.
Now we need an easy case in which (a + b + c) / d = integer doesn't imply abc/d = integer. I'd go for d = an odd prime. (For instance, a = 1, b = 4, c = 7, d = 3.)
S2::
Let's say a = n - 3, b = n - 2, c = n - 1, and d = n. Then we have
(n-3)(n-2)(n-1) / n
We can see that every term in the numerator will have an n in it, except the (-3)*(-2)*(-1), or -6. So the numerator will divide by n if and only if -6 divides by n. So the question becomes "is n a factor of 6?"
Further, we might notice that a, b, c, and d are all positive, and S2 implies that a < b < c < d, so d is at least 4. That means the ONLY possible value of d that's a factor of 6 is d = 6. So the question becomes "Is d = 6?" We can't answer yet, so this isn't sufficient.
S1 + S2::
Using our trick from S2 in conjunction with S1, we get (n - 3) + (n - 2) + (n - 1) is divisible by n, or (3n - 6) is divisible by n. Since n divides the first term (3n) it must also divide the second term (-6), so n is a factor of 6. We also know (from S2) that n is at least 4. Thus n MUST be 6. Hence a = 3, b = 4, c = 5, and d = 6, and we're done.
