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# Is x divisible by 30?

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Is x divisible by 30? [#permalink]  13 Mar 2008, 07:57
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-divisible-by-126857.html
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Re: DS - Divsibility [#permalink]  13 Mar 2008, 09:11
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B

$$30=2*3*5$$

1. $$x = k(m^3 - m)=k*m*(m^2-1)=k*m*(m-1)(m+1)$$

$$(m-1),m,(m+1)$$ are consecutive integers and the product of these integers will always be divisible by 6 (2 and 3).
But the consequence can be divisible by 5 or cannot be. Therefore, if k and the product are not divisible by 5, x will not be divisible by 30; if k or the product is divisible by 5, x will be divisible by 30. INSUFF.

2. $$x = n^5 - n=n*(n^4-1)=n*(n^2-1)*(n^2+1)=n*(n-1)*(n+1)*(n^2+1)$$

$$(n-1),n,(n+1)$$ are consecutive integers and the product of these integers will always be divisible by 6 (2 and 3).

if n=5k, 5k+1, or 5k+4, the product will be divisible by 5 and therefore, by 30.
if n=5k+2 or 5k+3 the product will not be divisible by 5.
but: n=5k+2: $$(n^2+1)=(5k+2)^2+1=25k^2+20k+5=5*(5k^2+4k+1)$$ Therefore, x is divisible by 30
n=5k+3: $$(n^2+1)=(5k+3)^2+1=25k^2+30k+10=5*(5k^2+6k+2)$$ Therefore, x is also divisible by 30.
SUFF.
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Re: DS - Divsibility [#permalink]  13 Mar 2008, 09:13
terp26 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

1: x = k(m^3 - m)
x = k m (m^2 - 1)
x = k m (m - 1) (m+1)
x = (k) (m - 1) (m) (m+1)

so x is a product of k and 3 consequtive integers (m-1) (m) and (m+1).
if k = 11 and m = 13, x is not divisible by 30.
if k = 10 and m = 13, x is divisible by 30. so nsf.

2: x = (n^5 - n) and integer n is >9,
x = n (n-1) (n+1) (n^2 + 1)

in this case, no matter the value of n, x is divisible by 30.
if n is, lets say, 12, x = 11(12) (13) (145) suff.
if n is, lets say, 13, x = 13 (12) (14) (13^2 + 1) = 12x13x14x140. suff.
if n is, lets say, 17, x = 16(17) (18) (290). so suff.

so B.
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Re: Is x divisible by 30? (1) x = k(m^3 - m), where m and k are [#permalink]  06 Feb 2014, 09:38
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Re: Is x divisible by 30? [#permalink]  07 Feb 2014, 04:09
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Is x divisible by 30?

(1) x = k*(m^3 - m), where m and k are both integers > 9 --> x=k*(m-1)m(m+1). Well, if k=30 then the answer is YES, but if k=11, then (m-1)m(m+1) only guarantees divisibility by 6, so the answer might be NO (for example m=12). Not sufficient.

(2) x = n^5 - n, where n is an integer > 9 --> the last digit of any integer in positive integer power repeats in pattern of 4 (some numbers have the pattern of 2 or 1, but pattern of 4 holds for them too). So, n^5 has the same last digit as n^1 or simply n, therefore the last digit of n^5-n is ALWAYS zero, so its ALWAYS divisible by 10. Next, n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=(n-1)n(n+1)(n^2+1) --> (n-1)n(n+1) is divisible by 3 (as there are 3 consecutive integers) --> n^5-n is divisible by 10*3=30. Sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-divisible-by-126857.html
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Re: Is x divisible by 30?   [#permalink] 07 Feb 2014, 04:09
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# Is x divisible by 30?

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