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Solving multiplications raised to monstrous powers?

Expert replies
by yates » Mon Jun 20, 2011 10:29 pm
This may be a ridiculously simple question, but I'll throw it out there.

On my GMATPowerPrep #1 test, I encountered a multiplication problem that consisted of two numbers -- both something like five or six digits long -- each raised by an incredibly high exponent, which consisted of six or six digits itself.

I won't state the exact question, because I plan to take the practice test again. But, based on what I've described, can anyone tell me how to solve this? General advice/instructions are fine (as responses from you); I'll figure it out from there.

Thanks!
GMATPrep1: (6/9/11); 690 (42 verbal; 41 quant); GMATPrep2: (6/22/11); 680 (42 verbal; 40 quant- ARGH!!)
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Source: — Problem Solving |

by kevincanspain » Mon Jun 20, 2011 10:39 pm
You may have noticed that the question was asking for the units digit or that the answer choices have different units digits. If so, you can think as follows:

What is the units digit of 177^177? (2^3= 2x2x2= 8)

When you square any integer whose units digit is 7 or 3, the result has a units digit of 9
When you square any integer whose units digit is 9, the result has a units digit of 1
When you square any integer whose units digit is 1, the result has a units digit of 1

Thus if k is a positive integer, 177^(4k) has a units digit of 1

177^177= 177^176 x 177
177^176 has a units digit of 1 since 176 is a multiple of 4
Therefore 177^177 has a units digit of 7.
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by Geva@EconomistGMAT » Mon Jun 20, 2011 10:45 pm
yates wrote:This may be a ridiculously simple question, but I'll throw it out there.

On my GMATPowerPrep #1 test, I encountered a multiplication problem that consisted of two numbers -- both something like five or six digits long -- each raised by an incredibly high exponent, which consisted of six or six digits itself.

I won't state the exact question, because I plan to take the practice test again. But, based on what I've described, can anyone tell me how to solve this? General advice/instructions are fine (as responses from you); I'll figure it out from there.

Thanks!
Obviously, you cannot be counted on to calculate the actual value of each of these powers, so there must be something else - some other aspect to the problem.

These can be:

1) Units digits
2) Break down the bases to prime factors, and combine like terms:

500^3* is actually (5*100)^3, which is equal to 5^3 * 100^3 = 125 * (10^2)3 = 125 * 10^6 = 125,000,000.

3) If the problem was actually an equation, the idea is to get the everything to the same bases on both sides, then ignore the bases and equate the exponents:

500^3 = 2^a * 5^b
5^3 * 10^3 = 2^a * 5^b
5^3 * 5^3 * 2^3 = 2^a * 5^b
5^6 * 2^3 = 2^a * 5^b

from which we can infer that a=3 and b=6.

There may be other cases, but I really need to look at the question. And so do you. Don't focus on what's obviously impossible to do (calculate the value) - focus on what the question is asking.
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by winniethepooh » Mon Jun 20, 2011 10:57 pm
@ kevincanspain: Hey, thanks for your help in solving this doubt, but what is finding units digit(Is it to find the first digit of the whole expression, and how did you get the answer to be 7?
Please answer, if possible!
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by kevincanspain » Tue Jun 21, 2011 5:54 am
The units digit of 123,456 is 6

The units digit of the product of two integers is the same as the units digit of the product of the units digit of those integers.


For example, the units digit of 734287 x 3834834 is 8 because the units digit of 7x4 is 8.
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by winniethepooh » Sat Jun 25, 2011 3:37 am
Thanks Kevin!
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