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Is integer Q evenly divisible by 6? (1) Q= 1, where N is a

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Is integer Q evenly divisible by 6? (1) Q= 1, where N is a [#permalink] New post 07 Oct 2003, 09:45
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Is integer Q evenly divisible by 6?

(1) Q=[7^N]–1, where N is a positive integer
(2) Q is evenly divisible by [2^2]*[7^7]*[3^3]
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 [#permalink] New post 07 Oct 2003, 12:20
D?

(1) Guessed the numbers and did some pattern matching.
(2) In other words Q is divisible by 108 * 7^7, since 108 is divisible by 6, so is Q.
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 [#permalink] New post 07 Oct 2003, 21:12
D is correct

(1) can you prove it?
(2) Q contains 2 and 3 and so divisible by 6.
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 [#permalink] New post 29 Nov 2003, 01:42
stolyar wrote:
D is correct

(1) can you prove it?
(2) Q contains 2 and 3 and so divisible by 6.


(1):

(x + y)^n will contains n + 1 terms in its expansion: n of those terms will have some factor of x and the last one will be y^n. Now let x = 6 and y = 1. Regardless of what n is, there will be one y^n or 1^n term in the expansion plus a whole bunch of other terms all divisible by 6. Hence, if we subtract out the y^n term, we are left with a sum of terms all divisible by x (or 6) hence the sum of those terms is also divisible by 6.
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 [#permalink] New post 29 Nov 2003, 02:33
(1)

x^n - y^n is always divisible by x-y.

Hence 7^7 - 1^7 is divisible by 7-1 = 6 :)

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 [#permalink] New post 29 Nov 2003, 20:02
Agree on D.


bhars18 wrote:
(1)

x^n - y^n is always divisible by x-y.

Hence 7^7 - 1^7 is divisible by 7-1 = 6 :)

Bharathi.

Good proof/method!
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 [#permalink] New post 29 Nov 2003, 21:49
wonder_gmat wrote:
Agree on D.


bhars18 wrote:
(1)

x^n - y^n is always divisible by x-y.

Hence 7^7 - 1^7 is divisible by 7-1 = 6 :)

Bharathi.

Good proof/method!


Hmm. Is this something that a GMAT candidate is expected to know? Can you derive this using elementary math?
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 [#permalink] New post 29 Nov 2003, 23:07
proof:

step 1.
n=1
x^n - y^n = (x-y)

step 2.
n=k
x^k-y^k = (x-y) * constant .........eq 1.

step 3.
n=k+1
x^(k+1) - y^(k+1) = x^(k+1) - y^(k+1) + xy^k - xy^k
=x(x^k-y^k) + y^k(x-y)
=x(x-y)*constant + y^k(x-y) ........from eq 1.
=(x-y)(x*constant+y^k)
=(x-y)*constant

:)
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 [#permalink] New post 30 Nov 2003, 00:51
dj wrote:
proof:

step 1.
n=1
x^n - y^n = (x-y)

step 2.
n=k
x^k-y^k = (x-y) * constant .........eq 1.

step 3.
n=k+1
x^(k+1) - y^(k+1) = x^(k+1) - y^(k+1) + xy^k - xy^k
=x(x^k-y^k) + y^k(x-y)
=x(x-y)*constant + y^k(x-y) ........from eq 1.
=(x-y)(x*constant+y^k)
=(x-y)*constant

:)


is that constant ALWAYS an integer??
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 [#permalink] New post 30 Nov 2003, 01:26
dj wrote:
proof:

step 1.
n=1
x^n - y^n = (x-y)

step 2.
n=k
x^k-y^k = (x-y) * constant .........eq 1.

step 3.
n=k+1
x^(k+1) - y^(k+1) = x^(k+1) - y^(k+1) + xy^k - xy^k
=x(x^k-y^k) + y^k(x-y)
=x(x-y)*constant + y^k(x-y) ........from eq 1.
=(x-y)(x*constant+y^k)
=(x-y)*constant

:)


Sorry. I don't see how you conclude step 2 from step 1. In fact, step 2 simply states what you are trying to prove.
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 [#permalink] New post 30 Nov 2003, 09:11
yes, step 3 is just the extrapolation of step 2. you can derive step 2 from step 1, as x^n - y^n = (x-y)*1 for n=1, x^k - y^k will follow the same pattern.

proof:

step 1.
n=1
x^n - y^n = (x-y)

step 2.
n=k
x^k-y^k = (x-y) * some interger .........eq 1. relate this with step 1.

step 3. .... extrapolates step 2.
n=k+1
x^(k+1) - y^(k+1) = x^(k+1) - y^(k+1) + xy^k - xy^k
=x(x^k-y^k) + y^k(x-y)
=x(x-y)*constant + y^k(x-y) ........from eq 1.
=(x-y)(x*constant+y^k)
=(x-y)*constant
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 [#permalink] New post 30 Nov 2003, 15:40
Can we not just apply few numbers from 1,2,3 for N and check for the result ?
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 [#permalink] New post 30 Nov 2003, 16:14
yes, the best way is to put in the integers and come onto the result. I was just trying to prove it using basic maths... ;)
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 [#permalink] New post 30 Nov 2003, 18:25
dj wrote:
yes, the best way is to put in the integers and come onto the result. I was just trying to prove it using basic maths... ;)


Yes, you can use numbers, but this does not prove that it is true for ALL n.

For example, it is not obvious to me why x^4132 - y^4132 MUST have (x-y) as one of its factors.

I'm sorry. It is not obvious to me why x^k - y^k = (x-y) * some number. In addition, step 3 says nothing different that step 2, but neither give me any rationale of why it is true. Your attempt at an inductive proof is incomplete - you must show 1) GIVEN x^k - y^k is divisible by x-y then IT FOLLOWS that x^(k+1) - y^(k+1) is also divisible by x-y. 2); it works for k = 1. Number 2 is fine, but number 1 doesn't work for me.

The explanation of why this is so is the essense of your "proof" and IMO you have simply hand-waved it by.
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 [#permalink] New post 30 Nov 2003, 21:01
when x and y are integers, (x+y) will be an integer.
for all the values of n, the result will be (x-y) * an integer.

n=2, (x-y)*(x+y)...this is an integer
n=3, (x-y)*(x^2+y^2+xy)...this is an integer
n=4, (x-y)(x+y)(x^2+y^2)....these 2 are integers
....

this leads to (x^n - y^n)/(x-y) = an integer.

think, this suffice.
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 [#permalink] New post 01 Dec 2003, 02:09
my initial idea:

(2) is obvious

(1) Q=[7^N]–1, where N is a positive integer
Q=[7^N]–[1^N], a distribution will always contain member [7–1], or 6.
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 [#permalink] New post 01 Dec 2003, 02:15
dj wrote:
when x and y are integers, (x+y) will be an integer.
for all the values of n, the result will be (x-y) * an integer.

n=2, (x-y)*(x+y)...this is an integer
n=3, (x-y)*(x^2+y^2+xy)...this is an integer
n=4, (x-y)(x+y)(x^2+y^2)....these 2 are integers
....

this leads to (x^n - y^n)/(x-y) = an integer.

think, this suffice.


All you have done is demonstrate that it works for the first 4 terms. While this certainly suggests the result, it is definitely not PROOF.
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Last edited by AkamaiBrah on 01 Dec 2003, 02:45, edited 1 time in total.
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 [#permalink] New post 01 Dec 2003, 02:26
stolyar wrote:
my initial idea:

(2) is obvious

(1) Q=[7^N]тАУ1, where N is a positive integer

Q=[7^N]тАУ[1^N], a distribution will always contain member [7тАУ1], or 6.


This certainly seems to be true but can you PROVE it for ALL n via a simple proof?

I.e.;Given Q = x^n - 1, prove that for ALL positive integers n, Q is a multiple of x-1

stolyar, i have faith in you.... 8-)
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 [#permalink] New post 30 Jan 2004, 02:45
dj wrote:
when x and y are integers, (x+y) will be an integer.
for all the values of n, the result will be (x-y) * an integer.

n=2, (x-y)*(x+y)...this is an integer
n=3, (x-y)*(x^2+y^2+xy)...this is an integer
n=4, (x-y)(x+y)(x^2+y^2)....these 2 are integers
....

this leads to (x^n - y^n)/(x-y) = an integer.

think, this suffice.


Relate the n-th step to (n+1)th. That is the trick of Mathematical Induction.
  [#permalink] 30 Jan 2004, 02:45
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