Units digit of n

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Units digit of n

by sreak1089 » Thu Aug 20, 2009 8:39 am
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

OA is A

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by niraj_a » Thu Aug 20, 2009 9:55 am
i got C

3 * 3 = 9

2) clearly insufficient as different values of y can yield different units valeus for n

1) x and y could be

1 and 6
2 and 5
3 and 4

3 * 3 * 1 * 6 = _ 4
3 * 3 * 2 * 5 = _ 0
3 * 3 * 3 * 4 = _ 8

so n could have either 4, 0, or 8 as units digits. insufficient.

1) and 2) together ensures the x and y combo to be 3 and 4, which means n will be have 8 as the units digit.

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by kaulnikhil » Thu Aug 20, 2009 11:09 am
cant be A 6,1 and 4,3 give us diff results ..should b C

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by pandeyvineet24 » Thu Aug 20, 2009 11:14 am
Yes should be C.

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by sreak1089 » Thu Aug 20, 2009 11:49 am
Surprisingly OA is A. I got this problem in MGMAT exam.

Explanation given is that since we need to find the last digit of (243)^x(463)^y, last digit can be found from 3^x.3^y = 3^ x+y). Since stmt # 1 gives x+y = 7, ans is A.

I am not convinced how (243)^x.(463)^y can evaluate to 3^x.3^y since 463 is not a multiple of 3...

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by pandeyvineet24 » Thu Aug 20, 2009 11:58 am
sreak1089 wrote:Surprisingly OA is A. I got this problem in MGMAT exam.

Explanation given is that since we need to find the last digit of (243)^x(463)^y, last digit can be found from 3^x.3^y = 3^ x+y). Since stmt # 1 gives x+y = 7, ans is A.

I am not convinced how (243)^x.(463)^y can evaluate to 3^x.3^y since 463 is not a multiple of 3...
ok, that means the question you initially posted had a small mistake. every one thought question to be
(243)* x * (463) * y
rather than (243)^x(463)^y.
Last edited by pandeyvineet24 on Thu Aug 20, 2009 12:00 pm, edited 1 time in total.

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by DanaJ » Thu Aug 20, 2009 11:59 am
Hey, my 1000th post!

Guys, the answer is A, but none of you are wrong. You just didn't see the question the way it was meant to be:

If (243^x)*(463^y) = n, where x and y are positive integers, what is the units digit of n?

The initial poster probably did not know how to write powers... Now, let's look at the problem and solve it. The last digit of the number n will be made up of the last digit of (243^x)*(463^y). Of course, we're only interested in the last digits of these two numbers, 243^x and 463^y. That means we're looking for the last digit of (3^x)*(3^y) = 3^(x+y), which is a power of 3.

As most of you know, last digits of powers of 3 repeat themselves, depending on the "format" of the power:
- if the power is divisible by 4, the LD = 1
- if the power has a remainder of 1 when divided by 4, the LD = 3
- if the power has a remainder of 2 when divided by 4, the LD = 9
if the power has a remainder of 3 when divided by 4, the LD = 7

A tells us that x + y = 7, and 7 divided by 4 yields a remainder of 3, which means that the LD = 7. A is sufficient.

B is insufficient since we don't have any info about y.