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## exponents problem

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revanth035 Just gettin' started!
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exponents problem Fri Mar 30, 2012 11:15 am
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whats the easiest solution to the problem (1.1)power 42

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Mike@Magoosh GMAT Instructor
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Fri Mar 30, 2012 2:50 pm
revanth035 wrote:
whats the easiest solution to the problem (1.1)power 42
Hi, there. I'm happy to help.

First of all, this is not a terribly likely question to be asked on the GMAT. Figuring something like this without a calculator is more advanced than what the GMAT typically covers.

The way to do it without a calculator is to use something called the Binomial Theorem, an advanced Precalculus topic. The Binomial Theorem say: when we raise the binomial (a + b) to the power of n, we get an expansion of (n+1) terms:
1) The powers of a go from n down to zero
2) The powers of b go from 0 up to n
3) The coefficients come from the nth row of Pascal's Triangle.

For example,
(a + b)^5 = 1(a^5) + 5(a^4)(b) + 10(a^3)(b^2) + 10(a^2)(b^3) + 5(a)(b^4) + 1(b^5)

where the numbers {1, 5, 10, 10, 5, 1} are the fifth row of Pascal's Triangle.

It's important to know that binomial coefficients, i.e. the numbers from Pascal's Triangle, can be calculated directly, without having to write out the whole triangle. In general, the nth row has (n+1)"places" from r = 0 to r = n. The numeral in the rth place of the nth row of Pascal's Triangle is given by
nCr = (n!)/[(r!)((n-r)!)]

(Incidentally, that is also the number of combinations of r things chosen from a pool of n. It's read, "n choose r.")

Now, to figure out an approximation of 1.1^42, think of it as (1 + 0.1)^42, where a = 1 and b = 0.1. Thus, successive terms will be getting smaller as the power of 1/10 increases, so we hope only the first few terms will contribute to a reasonable approximation.

(1 + 0.1)^42 = 1 + (42C1)*(0.1) + (42C2)*(0.01) + (42C2)*(0.001) + . . . .

42C1 = 42 (for any number n, nC1 = n) -----> (42C1)*(0.1) = 4.2

42C2 = (42*41)/(2*1) = 21*41 = 861 -----> (42C2)*(0.01) = 8.61

42C3 = (42*41*40)/(3*2*1) = 7*41*40 = 11480 -----> (42C2)*(0.001) = 11.48

Hmm, this is getting ugly without a calculator, and so far, the binomial coefficients are getting larger faster than the powers of ten in the denominators, so these terms are not getting smaller. Hmmm.

Cheating a bit, and checking with my calculator, the first 8 terms (up to r = 7) are greater than one, and we would have to calculate to the first 9 or 10 terms (r = 9) just to get an approximation to the nearest integer. Here are the first 10 terms:

1
4.2
8.61
11.48
11.193
8.50668
5.245786
2.6978328
1.18030185
0.44589181
0.1471442973

This is definitely not easy or short, but I know of no other non-calculator method of even beginning to get the answer.

FWIW, with a calculator, 1.1^42 = 54.76369924

With Wolfram Alpha, 1.1^42 = 54.763699237492901685126120802225273763666521

I feel very confident saying: the GMAT is not going to ask you to calculate the value of something like 1.1^42, since you are not allowed to have a calculator at all on the current GMAT. (When IR is introduced, there will be an onscreen calculator in the IR section only.)

It's really at the very outside of what the GMAT would even conceivably would ask, even something like 1.1^5. I would be a large tub of Ben & Jerry's that they never ever would ask 1.1^42.

That's my 2 cents. I hope what I said was helpful to you. Please let me know if you have any further questions.

Mike

_________________
Magoosh GMAT Instructor
http://gmat.magoosh.com/

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