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# If p is a prime number greater than 7, then (p^6)-1 is

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Manager
Joined: 28 Aug 2004
Posts: 205
If p is a prime number greater than 7, then (p^6)-1 is [#permalink]

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15 Oct 2004, 07:23
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If p is a prime number greater than 7, then (p^6)-1 is divisible by

1. 3*4*5
2. 4*5*6
3. 5*6*7
4. 7*8*9
5. 9*10*11
Senior Manager
Joined: 19 May 2004
Posts: 291

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15 Oct 2004, 08:04
Seems like choice 4 works.
Why ?!
Joined: 31 Dec 1969
Location: Russian Federation
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15 Oct 2004, 08:49
as it is a choice problem pick a value of p. 11 seems the most reasonable

11^6-1 = 1771560

this number is divisible by 7.
this number is divisible by 8 since 560 is divisible by 8
this number is divisible by 9 since 1+7+7+1+5+6+0=27 which is divisible by 9

implies this number is divisible by 7*8*9

PS:
For checking for divisibility by 7. There are many methods for checking
I use this one. From the number work from right to left multiply by 1,2,3,-1,-2,-3,1,2,3,.... and add the result

for this case 1*0 + 3*6 +2*5 - 1*1 - 3*7-2*7+1*1= -7 which is divisible by 7.

Also Dan have proposed a method in an earlier post.
Director
Joined: 20 Jul 2004
Posts: 592

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15 Oct 2004, 09:21
For that matter 1771560 is divisible by 3*4*5 too.
You have already proved 9 and 8. Last digit is 0, so 5 also is divisible.

What is the source of this Q?
Manager
Joined: 28 Aug 2004
Posts: 205

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15 Oct 2004, 11:25
I don't know what's the logic or pattern here, but for different primes, P^6 - 1 sometimes is and sometimes is not divisible by one or more of the choices, but is surely divisble by choice #4 (7*8*9 = 504). This is true for all primes above 7. It does not work for primes 2 or 3.

I have this question but not the OA. Maybe it involves algorithm and is beyond the GMAT scope (and maybe not) but am curious.

In the question stem, 7 is the lower limit, and it is also the lower limit in the 'correct' answer of 7*8*9.
Manager
Joined: 28 Aug 2004
Posts: 205

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16 Oct 2004, 15:54
ok... Let me give it a shot.

(p^6)-1 = P^3^2 - 1 = (P^3 - 1)(P^3 + 1)

Say n = number is 3, then 3^3 = 27 which is 1 less than a multiple of 7. Say n = 2, n^3 is 1 more than a multiple of 7.

therefore, for any number that is not a multiple of seven, then its cube will be one more or one less than a multiple of 7.

then all primes fit in (since primes have no multiples, not to mention 7), except 7 (since it is a multiple of 7), and this is why it is excluded in the question.

So, the divisor must be a multiple of 7, so answer 3 or 4, but since the question limits it for primes above 7, the other multiples have to be larger than 7 too.
16 Oct 2004, 15:54
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