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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] New post 27 Sep 2006, 14:14
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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
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Re: Hard divisibility DS [#permalink] New post 27 Sep 2006, 15:29
yezz 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


B
factors of 30 = 2*3*5

1
k*(m^3 - m) = k*m*(m^2-1) = k*m*(m+1)*(m-1)
any 3 consecutive integers > 9 will allow for divisibility by 2 and 3
but they may or maynot allow for for divisibility by 5
for eg,
m=21, k= 13 do not allow for divisibility by 5
m=23, k= 24 allows for divisibility by 5

insuffi

2
x = n^5 - n = n*(n^4-1) = n*(n^2+1)*(n+1)*(n-1)
any 3 consecutive integers > 9 will allow for divisibility by 2 and 3
but they may or maynot allow for for divisibility by 5,

the new expression here is (n^2+1)
for n= 11 , (n^2+1) = 122 is not divisible by 5, but 10=n-1 is
for n= 12 , (n^2+1) =145 is divisible by 5

suffi

Last edited by muddypaws on 27 Sep 2006, 18:29, edited 1 time in total.
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 [#permalink] New post 27 Sep 2006, 15:45
One more for E....good explanation muddypaws.
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 [#permalink] New post 27 Sep 2006, 17:57
Answer is B, the second condition is always satisfied for n > 9

and that is because of term n^2 + 1

since if the n leaves a remainder of 2 or 3 when divided by 5, then n-1 or n+1 is not divisible by 5 otherwise they are

If n leaves a remainder of 1 then n-1 is divisible by 5 and if n leaves remainder of 4 then n +1 is divisible by 5

for cases n leaves remainder of 2 and 3 the unit digit could be 2, 3, 7, 8

and n^2 + 1 will have last digit of (2X2) +1 , (3X3) +1, (7X7) +1, (8X8) +1

all of them divisible by 5

thus n^5 - n is divisible by 30
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 [#permalink] New post 27 Sep 2006, 18:30
jainan24 wrote:
Answer is B, the second condition is always satisfied for n > 9

and that is because of term n^2 + 1

since if the n leaves a remainder of 2 or 3 when divided by 5, then n-1 or n+1 is not divisible by 5 otherwise they are

If n leaves a remainder of 1 then n-1 is divisible by 5 and if n leaves remainder of 4 then n +1 is divisible by 5

for cases n leaves remainder of 2 and 3 the unit digit could be 2, 3, 7, 8

and n^2 + 1 will have last digit of (2X2) +1 , (3X3) +1, (7X7) +1, (8X8) +1

all of them divisible by 5

thus n^5 - n is divisible by 30


Good explanation. Indeed the answer should be (B)
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 [#permalink] New post 28 Sep 2006, 09:13
Opps sorry...perfect example for careless error.
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Re: Hard divisibility DS [#permalink] New post 29 Sep 2006, 00:07
yezz 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


statement 1: X= K*(m^3-m)

i.e X= K*(m-1)(m)(m+1).

Product of any n consecutive positive numbers is always divisible by n!.

So clearly (m-1)(m)(m+1) is divisible by 3! =6
Hence X is divisible by 6.
We are not sure whether x is divisible by 5 or not.
So 1 is not sufficient.

Statement 2: X= n^5-n
This will always end with 0.
So X is clearly divisible by 5.
X=n^5-n = n(n^4-1) = n[((n^2)^2) - 1^2]
ie X= n(n^2+1)(n^2-1)
X= n(n^2+1)(n+1)(n-1)
X= (n-1)(n)(n+1)(n^2+1)
Clearly X is divisible by 6.

So X must be divisibly by 30.

Hence B
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Last edited by cicerone on 25 Sep 2008, 00:46, edited 1 time in total.
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 [#permalink] New post 23 Oct 2006, 06:11
GMATT73 wrote:
jainan24 wrote:
Answer is B, the second condition is always satisfied for n > 9

and that is because of term n^2 + 1

since if the n leaves a remainder of 2 or 3 when divided by 5, then n-1 or n+1 is not divisible by 5 otherwise they are

If n leaves a remainder of 1 then n-1 is divisible by 5 and if n leaves remainder of 4 then n +1 is divisible by 5

for cases n leaves remainder of 2 and 3 the unit digit could be 2, 3, 7, 8

and n^2 + 1 will have last digit of (2X2) +1 , (3X3) +1, (7X7) +1, (8X8) +1

all of them divisible by 5

thus n^5 - n is divisible by 30


Good explanation. Indeed the answer should be (B)



Can please someone explain! I don't uderstand at all!
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Re: Hard divisibility DS [#permalink] New post 23 Oct 2006, 07:34
cicerone wrote:
yezz wrote:
Is x divisible by 30?

X= n(n^2+1)(n+1)(n-1)
X= (n-1)(n)(n+1)(n^2+1)
Clearly X is divisible by 6.

So X must be divisibly by 30.

Hence B


X= (n-1)(n)(n+1)(n^2+1)
Clearly X is divisible by 6.

So X must be divisibly by 30
.

Dude, You wana expand on 'X must be divisible by 30? I understand 3 consecutive numbers are always divisble by 6.
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 [#permalink] New post 23 Oct 2006, 08:21
N^5-n=n(n^2+1)(n^2-1)=n(n^2+1)(n+1)(n-1)
We already know n(n+1)(n-1) is divisible by 6, just need to know if the whole thing is divisible by 5.

if n=5k then it is divisible by 5.
if n=5k+1, n^2-1=25k^2+10k, divisible by 5.
if n=5k-1, n^2-1=25k^2-10k, divisible by 5.
if n=5k+2, n^2+1=25k^2+20k+5, divisible by 5.
if n=5k-2, n^2+1=25k^2-20k+5, divisible by 5.

I went the long way to find out that it is also divisible by 5. Cicerone, can you explain why n^5-n is always multiples of 10? Did you have to go through this long derivation or if there's a more straight forward way that we can see the result immediately?
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  [#permalink] 23 Oct 2006, 08:21
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