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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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27 Sep 2006, 15:14
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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]
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27 Sep 2006, 16:29
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^21) = k*m*(m+1)*(m1)
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^41) = n*(n^2+1)*(n+1)*(n1)
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=n1 is
for n= 12 , (n^2+1) =145 is divisible by 5
suffi
Last edited by muddypaws on 27 Sep 2006, 19:29, edited 1 time in total.



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One more for E....good explanation muddypaws.



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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 n1 or n+1 is not divisible by 5 otherwise they are
If n leaves a remainder of 1 then n1 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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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 n1 or n+1 is not divisible by 5 otherwise they are
If n leaves a remainder of 1 then n1 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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Opps sorry...perfect example for careless error.



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Re: Hard divisibility DS [#permalink]
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29 Sep 2006, 01:07
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^3m) i.e X= K*(m1)(m)(m+1). Product of any n consecutive positive numbers is always divisible by n!. So clearly (m1)(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^5n This will always end with 0. So X is clearly divisible by 5. X=n^5n = n(n^41) = n[((n^2)^2)  1^2] ie X= n(n^2+1)(n^21) X= n(n^2+1)(n+1)(n1) X= (n1)(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, 01:46, edited 1 time in total.



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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 n1 or n+1 is not divisible by 5 otherwise they are
If n leaves a remainder of 1 then n1 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]
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23 Oct 2006, 08:34
cicerone wrote: yezz wrote: Is x divisible by 30?
X= n(n^2+1)(n+1)(n1) X= (n1)(n)(n+1)(n^2+1) Clearly X is divisible by 6.
So X must be divisibly by 30.
Hence B
X= (n1)(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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N^5n=n(n^2+1)(n^21)=n(n^2+1)(n+1)(n1)
We already know n(n+1)(n1) 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^21=25k^2+10k, divisible by 5.
if n=5k1, n^21=25k^210k, divisible by 5.
if n=5k+2, n^2+1=25k^2+20k+5, divisible by 5.
if n=5k2, n^2+1=25k^220k+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^5n 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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