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If x is a positive integer, what is the remainder when

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If x is a positive integer, what is the remainder when [#permalink]

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New post 11 Aug 2008, 01:42
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If x is a positive integer, what is the remainder when \(7^{120x+5}+3\) is divided by 5?

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Re: PS: Divisible [#permalink]

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New post 11 Aug 2008, 02:39
hi,

the remainder should be 2.

i can write the equation as (7^120)^x*7^5+3.

Now, 7^5 will give the last digit as 7, and (7^120)^x will have 0 in its unit place.

The expression then can be read as (0+3)/5 which will leave the remainder 2.

Could someone please let me know if i am thinking correct.
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Re: PS: Divisible [#permalink]

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New post 11 Aug 2008, 02:56
siddarth wrote:
hi,

the remainder should be 2.

i can write the equation as (7^120)^x*7^5+3.

Now, 7^5 will give the last digit as 7, and (7^120)^x will have 0 in its unit place.

The expression then can be read as (0+3)/5 which will leave the remainder 2.

Could someone please let me know if i am thinking correct.



Completely wrong approach!!


Here's the way to solve the problem:

7^1/5 gives 2 as remainder.
7^2/5 gives 4 as remainder
:
7^4/5 gives 1 as remainder.

Thus 7^4n , where n can have any positive value, when divided by 5 will give 1 as the remainder.

Therefore, 7^120x/5 will give 1 as the remainder. (since 120 x is nothing but 4*30*x, i.e, 4*n ; where n= 30*x)

7^5/5 will give a remaider of 2.

Therefore, 7^120x +5 /5 +3 = 7^120x . 7^5/5 +3 = 2+3/5 = 0 as the remainder.



Thus the answer is 0.
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Re: PS: Divisible [#permalink]

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New post 11 Aug 2008, 03:07
lexis wrote:
If x is a positive integer, what is the remainder when \(7^{120x+5}+3\) is divided by 5?

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we know that x is a positive integer ... just put x = 1, we have to find last digit of 7^125

last digit of 7^1 = 7
last digit 0f 7^2 = 9
last digit of 7^3 = 3
last digit of 7^4 = 1
last digit of 7^5 = 7

so last digit of 7^125 = 7 ---- (125 = 4k+1)
last digit of 7^125 + 3 = 0
divided by 5 .... answer A
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Re: PS: Divisible [#permalink]

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New post 11 Aug 2008, 08:51
lexis wrote:
If x is a positive integer, what is the remainder when \(7^{120x+5}+3\) is divided by 5?

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\(7 ^{120x + 5}\) always mean that 5 will be last number of power of 7

Now powers of 7 repeat unit digit every 4 times i.e. 7 9 3 1 7

Since 120x is added to 5 it means dividing by four remainder will always be 1.

Therefore unit digit of \(7 ^{120x + 5}\) will always be 7 which added with 3 will give unit digit of 0 which will always be divisible by 5.

Therefore answer is 0
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Re: PS: Divisible [#permalink]

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New post 11 Aug 2008, 09:34
Your thinking is WAY TOO complex for the math that is required of the GMAT. These are not horribly hard questions. Each requires a unique approach which allows the question to be solved very easily. I've seen it said that if you can figure out the right approach to each question, the entire quant section can be completed in under 45 minutes and likely score a 50-51.

A big part of the GMAT is not so much our ability to pound through math problems, but our approach to the quesiton. Somoene that spends 7 minutes on a question and does multiple calculations is going to do worse than someone that sees the proper approach to the problem and gets it done in just over a minute. The one that took longer is going to get to fewer questions and therefore have to guess in the end.

siddarth wrote:
hi,

the remainder should be 2.

i can write the equation as (7^120)^x*7^5+3.

Now, 7^5 will give the last digit as 7, and (7^120)^x will have 0 in its unit place.

The expression then can be read as (0+3)/5 which will leave the remainder 2.

Could someone please let me know if i am thinking correct.

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Re: PS: Divisible [#permalink]

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New post 11 Aug 2008, 10:43
lexis wrote:
If x is a positive integer, what is the remainder when \(7^{120x+5}+3\) is divided by 5?

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7^1 7
7^2 49
7^3 ..3
7^4 ..1
7^5 ..7

same pattern continuous for every 5..
...7+3 = .......0 musb divisible by 5

answer should be 0
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Re: PS: Divisible [#permalink]

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New post 11 Aug 2008, 11:05
siddarth wrote:
durgesh79 wrote:
lexis wrote:
If x is a positive integer, what is the remainder when \(7^{120x+5}+3\) is divided by 5?

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1
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we know that x is a positive integer ... just put x = 1, we have to find last digit of 7^125

last digit of 7^1 = 7
last digit 0f 7^2 = 9
last digit of 7^3 = 3
last digit of 7^4 = 1
last digit of 7^5 = 7

so last digit of 7^125 = 7 ---- (125 = 4k+1)
last digit of 7^125 + 3 = 0
divided by 5 .... answer A



Got it thanks a lot.



This only works if we assume x=1, in which case m7^(120x+5)=m7^(125) and thus the units digit will be 7 as you mentioned. And if the units digit is 7, then 7+3 yields a 10, which when divided by 5 yields a 0.

But what happens when x is NOT equal to 1?

In that case, the last units digits will yield 9, 3, 1, 7, etc. which when added to 3 will yield, respectively, 12, 6, 4, and 10.

How is it then that 12, 6, or 4, divided by 5 will yield a 0?
m7^(120x+5)+3
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Re: PS: Divisible [#permalink]

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New post 11 Aug 2008, 11:23
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sarzan,

Regardless of the value of x, the units digit of 120x will always be zero. that units digit is going to be 0 * the units digit of x. 0 * any number = 0. So, if you add 5 to 0, you will always have 5.

I don't see how you arrived at what you are saying.

sarzan wrote:
This only works if we assume x=1, in which case m7^(120x+5)=m7^(125) and thus the units digit will be 7 as you mentioned. And if the units digit is 7, then 7+3 yields a 10, which when divided by 5 yields a 0.

But what happens when x is NOT equal to 1?

In that case, the last units digits will yield 9, 3, 1, 7, etc. which when added to 3 will yield, respectively, 12, 6, 4, and 10.

How is it then that 12, 6, or 4, divided by 5 will yield a 0?
m7^(120x+5)+3


lexis wrote:
If x is a positive integer, what is the remainder when \(7^{120x+5}+3\) is divided by 5?

0
1
2
3
4

_________________

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

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Re: PS: Divisible [#permalink]

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New post 11 Aug 2008, 11:44
jallenmorris wrote:
sarzan,

Regardless of the value of x, the units digit of 120x will always be zero. that units digit is going to be 0 * the units digit of x. 0 * any number = 0. So, if you add 5 to 0, you will always have 5.

I don't see how you arrived at what you are saying.

sarzan wrote:
This only works if we assume x=1, in which case m7^(120x+5)=m7^(125) and thus the units digit will be 7 as you mentioned. And if the units digit is 7, then 7+3 yields a 10, which when divided by 5 yields a 0.

But what happens when x is NOT equal to 1?

In that case, the last units digits will yield 9, 3, 1, 7, etc. which when added to 3 will yield, respectively, 12, 6, 4, and 10.

How is it then that 12, 6, or 4, divided by 5 will yield a 0?
m7^(120x+5)+3


lexis wrote:
If x is a positive integer, what is the remainder when \(7^{120x+5}+3\) is divided by 5?

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1
2
3
4



jallenmorris
Thanks! I can't even remember how I got to my answer! Nevermind. Kudos.
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Re: PS: Divisible [#permalink]

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New post 11 Aug 2008, 23:02
You are right! OA is A.
Re: PS: Divisible   [#permalink] 11 Aug 2008, 23:02
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