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# If 243^x*463^y = n, where x and y are positive integers

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If 243^x*463^y = n, where x and y are positive integers [#permalink]

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02 Oct 2010, 03:44
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If $$243^x*463^y =n$$ , where x and y are positive integers, what is the units digit of n?

(1) x + y = 7

(2) x = 4
[Reveal] Spoiler: OA

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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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02 Oct 2010, 03:56
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If $$243^x*463^y =n$$, where x and y are positive integers, what is the units digit of n?

The units digit of $$243^x$$ is the same as the units digit of $$3^x$$ and similarly the units digit of $$463^y$$ is the same as the units digit of $$3^y$$, so the units digit of $$243^x*463^y$$ equals to the units digit of $$3^x*3^y=3^{x+y}$$. So, knowing the value of $$x+y$$ is sufficient to determine the units digit of $$n$$.

(1) $$x + y = 7$$. Sufficient. (As cyclicity of units digit of $$3$$ in integer power is $$4$$, units digit of $$3^7$$ would be the same as of units digit of $$3^3$$ which is $$7$$)

(2) $$x=4$$. No info about $$y$$. Not sufficient.

Hope it helps.
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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03 Oct 2010, 08:50
Yup!! A it is....equation can be treated like 3^x*3^y hence (x+y)'s value can provide us the last digit...

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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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03 Mar 2014, 04:18
I have a doubt. Cyclicity of unit digit of 3 is 4. Hence we know that every fourth power of 3 (3^4, 3^8, 3^12) will have the same unit digit, 1. Hence when option B says x = 4, knowing that x and y are positive integers, we know that xy will be a multiple of 4. Unit digit of 3^4k is always 1 isn't it? Shouldn't this be sufficient information?

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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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03 Mar 2014, 04:21
siriusblack1106 wrote:
I have a doubt. Cyclicity of unit digit of 3 is 4. Hence we know that every fourth power of 3 (3^4, 3^8, 3^12) will have the same unit digit, 1. Hence when option B says x = 4, knowing that x and y are positive integers, we know that xy will be a multiple of 4. Unit digit of 3^4k is always 1 isn't it? Shouldn't this be sufficient information?

I think you are missing that $$3^x*3^y=3^{x+y}$$, so the exponent is x+y not xy.

Does this make sense?
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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03 Mar 2014, 04:26
Yes! Can't believe I just made that mistake. Such mistakes are gonna cost me. :/

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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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03 Mar 2014, 04:28
siriusblack1106 wrote:
Yes! Can't believe I just made that mistake. Such mistakes are gonna cost me. :/

Yes, careless errors are the #1 cause of score drops on the GMAT! They cause you to miss easier questions, hurting your score a lot more than not know how to solve the harder ones. So, be more careful, before you submit your answer, double-check that it’s the answer to the proper question.
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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18 May 2014, 08:53
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243 = 3^5

463 ends with a 3. So we have to know how many times we will multiply 3's at the end of each numbers.

1) 7 times - SUF
2) we dont know Y - INSUF

Choose (a)

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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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02 Nov 2014, 17:48
Bunuel wrote:
If $$243^x*463^y =n$$, where x and y are positive integers, what is the units digit of n?

The units digit of $$243^x$$ is the same as the units digit of $$3^x$$ and similarly the units digit of $$463^y$$ is the same as the units digit of $$3^y$$, so the units digit of $$243^x*463^y$$ equals to the units digit of $$3^x*3^y=3^{x+y}$$. So, knowing the value of $$x+y$$ is sufficient to determine the units digit of $$n$$.

(1) $$x + y = 7$$. Sufficient. (As cyclicity of units digit of $$3$$ in integer power is $$4$$, units digit of $$3^7$$ would be the same as of units digit of $$3^3$$ which is $$7$$)

(2) $$x=4$$. No info about $$y$$. Not sufficient.

Hope it helps.

Hi Bunuel,

But surely shouldn't it matter if 3 is raised to 1 and or 6? Meaning, if it's 3^3 + 3^4 = 7 + 1 = 8. But, if its 3^2+3^5 = 9 + 3 = 12, units of 2. Doesn't that yield insufficient?

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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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03 Nov 2014, 00:50
russ9 wrote:
Bunuel wrote:
If $$243^x*463^y =n$$, where x and y are positive integers, what is the units digit of n?

The units digit of $$243^x$$ is the same as the units digit of $$3^x$$ and similarly the units digit of $$463^y$$ is the same as the units digit of $$3^y$$, so the units digit of $$243^x*463^y$$ equals to the units digit of $$3^x*3^y=3^{x+y}$$. So, knowing the value of $$x+y$$ is sufficient to determine the units digit of $$n$$.

(1) $$x + y = 7$$. Sufficient. (As cyclicity of units digit of $$3$$ in integer power is $$4$$, units digit of $$3^7$$ would be the same as of units digit of $$3^3$$ which is $$7$$)

(2) $$x=4$$. No info about $$y$$. Not sufficient.

Hope it helps.

Hi Bunuel,

But surely shouldn't it matter if 3 is raised to 1 and or 6? Meaning, if it's 3^3 + 3^4 = 7 + 1 = 8. But, if its 3^2+3^5 = 9 + 3 = 12, units of 2. Doesn't that yield insufficient?

Are you sure you are reading the question correctly? It's 243^x*463^y, 243^x multiplied by 463^y not 243^x + 463^y...
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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28 Jul 2016, 14:01
Well I don't agree the answer should be indeed D.
Option 1 suggests x+y=7 this can have multiple x and y combinations like (1,6) (2,5) (4,3) and so on so the units digit of 243^x and 463^y will differ .

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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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28 Jul 2016, 17:37
sandeep211986 wrote:
Well I don't agree the answer should be indeed D.
Option 1 suggests x+y=7 this can have multiple x and y combinations like (1,6) (2,5) (4,3) and so on so the units digit of 243^x and 463^y will differ .

Hey Buddy,

All the combinations, for x+y=7, will yield the same units digit. Consider the following
x=1,y=6
243*463*463*463*463*463*463 ~~ To find units digit we just need 3^1 * 3^6 = 3^7, i.e 7 (units digit of 2187)

Same goes with other combinations. 3^7 ends up deciding the units digit.

Should the question would have been something like, 245^x * 463^y = n, the combinations of different values of x & y would have yielded different units digits.

Hope that clears

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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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20 Dec 2016, 14:25
amandeep_k wrote:
If (243) x(463) y = n, where x and y are positive integers, what is the units digit of n?

(1) x + y = 7

(2) x = 4

I can't get how the answer can be A.

Had it been $$243^x * 463^y$$. In this case we'll get same base 3 and we can add the powers

$$3^0*3^7$$
$$3^1*3^6$$
$$3^2*3^5$$
...

In each case we'll get $$3^7$$ which units digit we can identify. That that will be sufficient.

But we have 3*x*3*y = 9*x*y which can take any values. Not sufficient. I can't get the idea how answer can be A.

Please correct me if I'm wrong.

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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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20 Dec 2016, 14:39
vitaliyGMAT wrote:
amandeep_k wrote:
If (243) x(463) y = n, where x and y are positive integers, what is the units digit of n?

(1) x + y = 7

(2) x = 4

I can't get how the answer can be A.

Had it been $$243^x * 463^y$$. In this case we'll get same base 3 and we can add the powers

$$3^0*3^7$$
$$3^1*3^6$$
$$3^2*3^5$$
...

In each case we'll get $$3^7$$ which units digit we can identify. That that will be sufficient.

But we have 3*x*3*y = 9*x*y which can take any values. Not sufficient. I can't get the idea how answer can be A.

Please correct me if I'm wrong.

Hi, You are right. my question was wrong. Sorry for the inconvenience.

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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]

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18 Dec 2017, 01:58
A.
If we map the cyclicity of 3--> with x + y values --> 3 and 4/ 4 and 3/ 1 and 6; 6 and 1/ 5 and 2; 2 and 5 [3,9,7,1,3,9,7,1...] --> the answer is always 7. [7*1 = 7; 3*9 = _7..so on]

St 2 . No value for y

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Re: If 243^x*463^y = n, where x and y are positive integers   [#permalink] 18 Dec 2017, 01:58
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