Author 
Message 
Manager
Joined: 28 Aug 2004
Posts: 205

If p is a prime number greater than 7, then (p^6)1 is [#permalink]
Show Tags
15 Oct 2004, 07:23
1
This post was BOOKMARKED
This topic is locked. If you want to discuss this question please repost it in the respective forum.
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

Seems like choice 4 works.
Why ?!



Joined: 31 Dec 1969
Location: Russian Federation
Concentration: Entrepreneurship, International Business
GMAT 3: 740 Q40 V50 GMAT 4: 700 Q48 V38 GMAT 5: 710 Q45 V41 GMAT 6: 680 Q47 V36 GMAT 9: 740 Q49 V42 GMAT 11: 500 Q47 V33 GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)

as it is a choice problem pick a value of p. 11 seems the most reasonable
11^61 = 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*72*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

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

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

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.










