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If 243^x*463^y = n, where x and y are positive integers
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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
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Re: If 243^x*463^y = n, where x and y are positive integers
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02 Oct 2010, 03:56
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. Answer: A. Hope it helps.
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Re: If 243^x*463^y = n, where x and y are positive integers
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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
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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?
Shouldn't the answer be D?



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Re: If 243^x*463^y = n, where x and y are positive integers
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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?
Shouldn't the answer be D? 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
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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
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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, doublecheck 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
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18 May 2014, 08:53
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
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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.
Answer: A.
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
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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.
Answer: A.
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
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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
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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 24 3*46 3*46 3*46 3*46 3*46 3*46 3 ~~ 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
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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
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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
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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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If 243^x*463^y = n, where x and y are positive integers
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13 May 2020, 19:40
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.
Answer: A.
Hope it helps. I don't think you can do this \(3^x*3^y=3^{x+y}\) Because of the bases i.e. 243 ≠ 463 Can you explain please why did you do this? Thanks!



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Re: If 243^x*463^y = n, where x and y are positive integers
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13 May 2020, 23:01
D4kshGargas 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.
Answer: A.
Hope it helps. I don't think you can do this \(3^x*3^y=3^{x+y}\) Because of the bases i.e. 243 ≠ 463 Can you explain please why did you do this? Thanks! I think I explained this in the highlighted part. No?
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Re: If 243^x*463^y = n, where x and y are positive integers
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14 May 2020, 19:27
Bunuel wrote: D4kshGargas 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.
Answer: A.
Hope it helps. I don't think you can do this \(3^x*3^y=3^{x+y}\) Because of the bases i.e. 243 ≠ 463 Can you explain please why did you do this? Thanks! I think I explained this in the highlighted part. No? Yeah, but I couldn't intuitively understand it back then. Thanks tho



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Re: If 243^x*463^y = n, where x and y are positive integers
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01 Jun 2020, 08:55
Bunuel wrote: D4kshGargas 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.
Answer: A.
Hope it helps. I don't think you can do this \(3^x*3^y=3^{x+y}\) Because of the bases i.e. 243 ≠ 463 Can you explain please why did you do this? Thanks! I think I explained this in the highlighted part. No? I'm not sure this is right x+y = 7 will give you 3 different combinations of x and y to add up to 7: 1&6 2&5 3&4 With cyclicality of 4 you can get the same answer for the first two pairs, but you don't get the same answer for the third pair. cyclicity for 3 is: 3971 remainder for 1 & 6 are follows : 3 + 9 remainder for 2 & 5 are follows : 9 + 3 but remainder for the third is different: remainder for 3 & 4 are follows : 7 + 1



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Re: If 243^x*463^y = n, where x and y are positive integers
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01 Jun 2020, 09:12
Andrewcoleman wrote: Bunuel wrote: D4kshGargas 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.
Answer: A.
Hope it helps. I don't think you can do this \(3^x*3^y=3^{x+y}\) Because of the bases i.e. 243 ≠ 463 Can you explain please why did you do this? Thanks! I think I explained this in the highlighted part. No? I'm not sure this is right x+y = 7 will give you 3 different combinations of x and y to add up to 7: 1&6 2&5 3&4 With cyclicality of 4 you can get the same answer for the first two pairs, but you don't get the same answer for the third pair. cyclicity for 3 is: 3971 remainder for 1 & 6 are follows : 3 + 9 remainder for 2 & 5 are follows : 9 + 3 but remainder for the third is different: remainder for 3 & 4 are follows : 7 + 1 I'm not sure I understand what you mean. Are you saying that the units digit of 3^(3+4) is not the same as the units digit of 3^3*3^4? If so, then since 3^(3+4) = 3^3*3^4, the untis digit are the same (both equal to 2187).
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Re: If 243^x*463^y = n, where x and y are positive integers
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