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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]

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New post 24 Jan 2012, 09:14
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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]

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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]

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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]

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

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New post 31 Jan 2018, 05:03
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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


I solved it this way..

\({11^{23}}*{3^{46}}\) = \((100-1)^{99}\)
As the first term of the bracket is 100.. In the expansion of the above expression, we will get a number which will have minimum of 2 Zeros at the end. So, its basically
\(100-1 = 99\)
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Re: What is the last two digits of 11^(23)*3^(46)? [#permalink]

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New post 31 Jan 2018, 10:09
rahul16singh28 wrote:
LM wrote:
What is the last two digits of \({11^{23}}*{3^{46}}\)?

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


I solved it this way..

\({11^{23}}*{3^{46}}\) = \((100-1)^{99}\)
As the first term of the bracket is 100.. In the expansion of the above expression, we will get a number which will have minimum of 2 Zeros at the end. So, its basically
\(100-1 = 99\)


Absolute....
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Re: What is the last two digits of 11^(23)*3^(46)? [#permalink]

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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


Notice that 11^23 x 3^46 = 11^23 x (3^2)^23 = 11^23 x 9^23 = 99^23

Notice that 99 is 1 less than 100; thus, when we raise 99 to the 23rd power, we can consider it as raising “-1” to the 23rd power. Since (-1)^23 = -1, it means we can obtain the last two digits by subtracting 1 from 100. Thus the last two digits must be 99.

Even if we don’t know that determining the last two digits of raising 99 to a positive integer power is “same” as raising -1 to a positive integer power, we can just check the last two digits of 99 by raising it to some small positive integer powers and determining if there is a pattern.

99^1 = 99

99^2 = 9,801

99^3 = 970,299

99^4 = 96,059,601

We see that when 99 is raised to an odd power, the last two digits are 99 and when it is raised to an even power, the last two digits are 01. Since the power of 23 is odd, then the last two digits must be 99.

Answer: E
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Re: What is the last two digits of 11^(23)*3^(46)?   [#permalink] 02 Feb 2018, 11:02
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