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What is the last two digits of 11^(23)*3^(46)?

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What is the last two digits of 11^(23)*3^(46)? [#permalink] New post 24 Jan 2012, 09:14
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Question Stats:

55% (03:07) correct 45% (02:16) wrong based on 151 sessions
What is the last two digits of \({11^{23}}*{3^{46}}\)?

A. 01
B. 09
C. 29
D. 49
E. 99
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Jan 2012, 11:09, edited 2 times in total.
Edited the OA, should be E not C
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Re: What is last two digit of [#permalink] New post 24 Jan 2012, 11:06
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LM wrote:
What is the last two digits of \({11^{23}}*{3^{46}}\)?

A. 01
B. 09
C. 29
D. 49
E. 99


\(11^{23}*3^{46}=11^{23}*9^{23}=99^{23}\) --> the last two digits of 99 in odd power is 99 (for example 99^1=99) and the last two digits of 99 in positive even power is 01 (for example 99^2=9801). Hence the last two digits of \(99^{23}\) will be 99.

Answer: E.

P.S. You won't need this for the GMAT.
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Re: What is last two digit of [#permalink] New post 23 Aug 2012, 18:29
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Bunuel wrote:
LM wrote:
What is the last two digits of \({11^{23}}*{3^{46}}\)?

A. 01
B. 09
C. 29
D. 49
E. 99


\(11^{23}*3^{46}=11^{23}*9^{23}=99^{23}\) --> the last two digits of 99 in odd power is 99 (for example 99^1=99) and the last two digits of 99 in positive even power is 01 (for example 99^2=9801). Hence the last two digits of \(99^{23}\) will be 99.

Answer: E.

P.S. You won't need this for the GMAT.


Here is another way that I give credit to Bunuel. What is the last two digits of \({11^{23}}*{3^{46}}\)? is the same question as what is the remainder of \({11^{23}}*{3^{46}}\) when divided by 100. Because its remainder is the same as the last two digit.

Ex. 152/10=15.2 (where the remained is 2=(.2)*10 )
269/100=2.69 (where the remained is 69=(.69)*100)

Now here is Bunuel's really cool trick:

\(11^{23}*3^{46}=11^{23}*9^{23}=99^{23}\)

Now to find the remainder of \(99^{23}\) to any number (in this case its a 100) we break up the number into one part that is a multiple of the number and the other a remainder.
So \(99^{23}=(100-1)^{23}\).

We take out remainder -1 and we find its remainder to the number (in this case its a 100)

-1=100(0)-1 (now because we can't have a negative remainder we add 100 to the remainder on the right handside to be remainder of r=99)
(-1)^2=1=100(0)+1 (remainder doesn't need adjustment, r=1)
(-1)^3=100(0)-1 (r=99)
(-1)^4=1=100(0)+1 (r=1)

We can see the periodicity here, r=99 for odd powers and r=1 for even powers.
Since our exponent here is even, the remainder is r=99.

I'll do another example so others can get the concept I laid above.

So lets say we want to find the remainder of 52^21 when divided by 7.

So Bunuel's hat trick is to break the number (52)^21=(2*7+3)^21

Now 3 is the left over portion. So the question becomes what is the remainder 3^21 when divided by 7. To find this we have to find the periodicities.

3=7(0)+3 (r=3)
3^2=9=7(1)+2 (r=2)
3^3=27=7(3)+6 (r=6)
3^4=81=7(11)+4 (r=4)
3^5=243=7(34)+5 (r=5)
3^6=729=7(104)+1 (r=1)
3^7=2187=7(3012)+3 (r=3)

We can stop once we see the remainders begin repeating after the sixth iteration. Since our power is ^21 that puts us on 21=6(3)+3 on the third power, so the remainder will r=6. Pretty neat, huh? Bunuel's hat trick, Ha! We shall it call it so from now on. I picked numbers that had very long periods( here is 6 ). Question is their a shortcut to figure out how many periods without, actually plugging and chugging until the remainders repeat?

For example: whats the remainder of (32)^11 when divided by 7?
Bunuels hat trick (32)^11=(28+4)^21

4=7(0)+4
4^2=7(2)+2
4^3=7(9)+1
4^4=7(36)+4

Here we have a period of 3. Anyone know? Thank you.
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Re: What is the last two digits of 11^(23)*3^(46)? [#permalink] New post 16 Feb 2014, 14:27
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Re: What is the last two digits of 11^(23)*3^(46)? [#permalink] New post 15 Mar 2015, 22:58
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Re: What is the last two digits of 11^(23)*3^(46)? [#permalink] New post 09 Jun 2015, 15:38
Bunuel - Your explanation is really elegant. But can you help me understand as to why this isn't needed for GMAT
Re: What is the last two digits of 11^(23)*3^(46)?   [#permalink] 09 Jun 2015, 15:38
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