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

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05 Nov 2010, 04:36

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mattapraveen wrote:

If [(243)^x]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

The above question is from MGMAT. The answer given is A.

But i believe, even (X=4) will also give the solution.

As X=4, the power of 3 in the expression [(243)^x]X[(463)^y] would always be 3^4n and the units digit will always be 1.

Can some one comment on this. Bunuel your help would be very much appreciated.

The units digit of \(243^x\) equals to the units digit of \(3^x\) and the units digit of \(463^y\) equals to the units digit of \(3^y\) (general rule). Hence 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 \(3\) is \(4\), units digit of \(3^7\) would be the same as of units the digit of \(3^3\) which is \(7\))

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

For example if \(y=1\) then the units digit of \(3^{x+y}=3^5\) will be 3, but if \(y=4\) then the units digit of \(3^{x+y}=3^8\) will be 1.

If [(243)^x]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

The above question is from MGMAT. The answer given is A.

But i believe, even (X=4) will also give the solution.

As X=4, the power of 3 in the expression [(243)^x]X[(463)^y] would always be 3^4n and the units digit will always be 1.

Can some one comment on this. Bunuel your help would be very much appreciated.

Let me try ... i am not even 1% what bunuel is ..

The reason why x=4 is insufficient because we dont know anything about Y, so if x=4 and y is 1, the unit digit will be 3, if y=2 then, unit digit will be 9 and so on.

So, i think it is necessary to have exact value of both the variants before reaching to the unit digit.

Had it been 5 as base , then it would not have made any difference because the unit digit is anyhow gonna be 5 only.

am open to comments .. correct me if i am wrong anywhere

Units digit of 243^x equals to units digit of 3^x and units digit of 463^y equals to units digit of 3^y (general rule). Hence units digit of 243^x*463^y equals to units digit of 3^x*3^y=3^{x+y}. So knowing the value of x+y is sufficient to determine units digit of n.

(1) x+y=7. Sufficient. (As cyclicity of 3 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.

For example if y=1 then units digit of 3^{x+y}=3^5 will be 3, but if y=4 then units digit of 3^{x+y}=3^8 will be 1.

Answer: A.

For more on this issue check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.

Thanks Bunuel. Yeah, got it. Some how got into an illusion that (X+Y)=(X*Y) . Hence was assuming when X=4, Y will be 4*Y.

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

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31 Jul 2014, 09:39

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