Is x divisible by 30? (1) X = k*(m^3 - m), where m and k are : GMAT Data Sufficiency (DS)
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# Is x divisible by 30? (1) X = k*(m^3 - m), where m and k are

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Is x divisible by 30? (1) X = k*(m^3 - m), where m and k are [#permalink]

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17 Jul 2003, 02:14
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Is x divisible by 30?

(1) X = k*(m^3 - m), where m and k are both integers > 9
(2) X = n^5 - n, where n is an integer > 9

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
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17 Jul 2003, 06:32
B

you take any number where n>9 you and subtract that by 1 you end up in the last integer being either 5 or 0, which will certainly be divided 30.
I hope I am right?

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17 Jul 2003, 09:25
(1) X=k(m^3-m)=km(m^2-1)=km(m-1)(m+1)

the part m(m-1)(m+1) is always divisible by 3, since those are three consecutive numbers. Since m>9, let us check some numbers
m=10) 9*10*11 - OK
m=11) 10*11*12 - OK
m=12) 11*12*13 - not OK. but we have K>9 can be 11 - not OK again

(2) X=n^5-n=n(n^4-1)=n(n-1)(n+1)(n^2+1)

n(n-1)(n+1) is divisible by 3, but not by 30 each time

consider n=13) 12*13*14*170 - OK
consider n=12) 11*12*13*145 - divisible by 30
consider n=11) 10*11*12*122 - divisible by 30
consider n=10) 9*10*11*82 - divisible by 30

Looks like B.
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17 Jul 2003, 14:54
Cool Stolyar. Just for fun, can you quickly prove that (2) is ALWAYS divisible by 30 for ALL n?
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18 Jul 2003, 00:39
satgates wrote:
B

you take any number where n>9 you and subtract that by 1 you end up in the last integer being either 5 or 0, which will certainly be divided 30.
I hope I am right?

The answer is indeed B, but I don't how you can show mathematically that (2) always has last digit 0 or 5 (actually, it always has last digit 0). I would be interested as that would be a great quick proof (although, my proof indirectly implies that the last digit must be 0).
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18 Jul 2003, 05:02
AkamaiBrah wrote:
Cool Stolyar. Just for fun, can you quickly prove that (2) is ALWAYS divisible by 30 for ALL n?

cannot prove that

X=n(n-1)(n+1)(n^2+1), where n>9

n(n-1)(n+1) is aways divisible by 6
but (n^2+1) is not always divisible by 5

n(n-1)(n+1)(n^2+1)

if n is odd then each of (n-1), (n+1), (n^2+1) is even
if n is even then each of (n-1), (n+1), (n^2+1) is odd

I do not know what to do next...
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18 Jul 2003, 07:41
Quote:
Is x divisible by 30?

(1) X = k*(m^3 - m), where m and k are both integers > 9
(2) X = n^5 - n, where n is an integer > 9

Solution:

First, since 30 = 2*3*5, what the question stem is REALLY asking you is:
Does X have at least one factor of 2, 3, and 5?

(1). When you see an expression that you cannot make any conclusions about, immediate think about how to rearrange it!

for m^3 * m , the only thing you can do is factor. So:

k(m^3 - m) = k(m(m^2 - 1)) = k*(m-1)*m*(m+1) = k X product of 3 consecutive numbers.

Since every third number must be a factor of 3, and every two numbers must be even, (m-1)*m*(m+1) MUST have a factor of 6. However, there is no way to determine whether it has a factor of 5, nor can we determine this about k, hence, (1) is NOT sufficient. This analysis should only have taken 30 seconds. (Don't plug in numbers -- this is plenty proof - move on!).

(2). Right away, factor until you can't anymore!

n^5 - n = n * (n^4 - 1)
= n * (n^2 - 1) * (n^2 + 1)
= n * (n - 1) * (n + 1) * (n^2 + 1)
= (n - 1) * n * (n + 1) * (n^2 + 1) or 3 consec integers x (n^2 + 1)
(Stolyar, you got this far fine!!!)

We already know that the 3 consecutive integers MUST have a factor of 6. So we need to determine whether their is a factor of 5 in there somewhere for all possible n. Use logic.

Okay, n must either be divisible by 5, or have remainder of 1, 2, 3, or 4, when divided by 5, right?

If n is divisible by 5, we have our 5, YES!
If n has remainder 1, then the (n - 1) term must be divisible by 5, YES again!
if n has remainder 4, then the (n + 1) term must be divisible by 5, YES again!

Now we are only left with when n has remainder 2 and 3 respectively. For this, we need to check the (n^2 + 1) term

case 1: n divided by 5 has remainder 2.
we can express n as
n = 5j + 2.
so n^2 + 1 = (5j + 2)^2 + 1
= 25j^2 + 20j + 4 + 1
= 25j^2 + 20j + 5 --> YES

case 2: n divided by 5 has remainder 3.
we can express n as
n = 5j + 3.
so n^2 + 1 = (5j + 3)^2 + 1
= 25j^2 + 30j + 9 + 1
= 25j^2 + 30j + 10 --> YES

Hence X is divisible by 5 for all possible n's so (2) is sufficient and the answer is (b).

Author's note: the reason that I have a K term in (1) is so that there exists an X that satisfies both equations. NOBODY ever points this out, but if you look carefully at all of the ETS problems in the OG, you will not be able to (at least I couldn't) find one where (1) and (2) contradict each other, even though they have never officially admitted that this is always true. This is good to know, because if you find that that they DO contradict, I believe that you probably made an error. (Most of the prep courses are not so careful). Also, the reason I specified all variables > 9 is to dissuade people from plugging in numbers -- much faster to use logic and number properties.

BTW, if K = 101, m = 10, and n = 10, X = 9990.
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MBA, Anderson School of Management, UCLA, Class of 1993

Last edited by AkamaiBrah on 18 Jul 2003, 07:50, edited 1 time in total.
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18 Jul 2003, 07:49
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18 Jul 2003, 16:56
AkamaiBrah wrote:
Cool Stolyar. Just for fun, can you quickly prove that (2) is ALWAYS divisible by 30 for ALL n?

n(n-1)(n+1)(n^2+1) is divisible by 2 and 3. What's left is to prove that it is divisible by 5.

So, If i can prove that the difference of

(n+1)[(n+1) - 1] [(n+1) + 1] [(n+1)^2 + 1] - n(n-1)(n+1)(n^2+1) is divisible by 5 then the number is divisible by 5.

So going by that i have,

n(n+1)[(n+2)(n^2+2n+2) - (n-1)(n^2+1)]

that is, n(n+1)[n^3+4n^2+6n+4 - n^3 + n^2 + n -1]

that is n(n+1)[ 5n^2 + 5n + 5]

that is 5n(n+1)(n^2+n+1)

So the number above is divisible by 5.

So it means that n^5 - n is divisible by 30 for ALL N.
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04 Nov 2010, 08:56
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you can also do this quick trick every time they ask you for divisions of some sums...for example we can write the following
n^2 + 1 = n^2 + 5 - 4 = n^2-2^2 + 5 = (n-2)(n+2) + 5 ...you can develop the equation to have all terms divide 5 including this one:
(n-2)(n+2)(n-1) ...for n>9 ...
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04 Nov 2010, 14:20
AkamaiBrah wrote:
Is x divisible by 30?

(1) X = k*(m^3 - m), where m and k are both integers > 9
(2) X = n^5 - n, where n is an integer > 9

I'm sure most people here will have worked on problems involving units digits and powers - something which is frequently tested on the GMAT - questions like What is the units digit of $$3^{15} + 4$$? . In solving such questions, you pretty much always make use of the fact that units digits 'repeat in blocks of four'. For example, for powers of 3:

3^1 ends in 3
3^2 ends in 9
3^3 ends in 7
3^4 ends in 1
3^5 ends in 3, and we're back to where we started; 3^5 has the same units digit as 3^1

That happens for any digit - units digits repeat in blocks of 4 (well, that's the worst case; sometimes we have blocks of 2 or of 1). But for any integer k, it is always true that k^5 has the same units digit as k, and similarly, k^6 has the same units digit as k^2, and so on. As a consequence, if k is an integer, k^5 - k always ends in 0, so is always divisible by 10 (and by the same token, you can invent all kinds of expressions which must be multiples of 10; k^9 - k, for example, or k^102 - k^2) .

So here, from Statement 2, n^5 - n must be divisible by 10, and since n^5 - n = (n-1)(n)(n+1)(n^2 + 1) contains a product of three consecutive integers, it must be divisible by 3 as well, and thus must be divisible by 30, so Statement 2 is sufficient.

Oh, and the answer to the question above, What is the units digit of $$3^{15} + 4$$? , is 1, since 3^15 has the same units digit as 3^3, so 3^15 ends in 7. When we add 4, we get a number ending in 1.
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Re: DS number theory   [#permalink] 04 Nov 2010, 14:20
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