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.