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# Easier way to find powers

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Manager
Joined: 22 Mar 2011
Posts: 65

Kudos [?]: 16 [0], given: 10

Easier way to find powers [#permalink]

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20 Nov 2011, 22:02
Hello all

I am here to ask for a quick way to find results of numbers with (large) powers.

In one GMAT prep Q

2^5+2^5+3^5+3^5+3^5 can be solved as

2^5(1+1) + 3^5 (1+1+1)
2^6+3^6

But what if we have larger powers
e.g.
3^180 - 3^30 or
3^180 - 3^30 + 2 ^120 - 2^60

What is an easy way to solve such problems with large powers when base has addition and sub.

thanks!

Kudos [?]: 16 [0], given: 10

Senior Manager
Joined: 23 Jun 2009
Posts: 360

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Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Re: Easier way to find powers [#permalink]

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21 Nov 2011, 02:07
melguy wrote:
Hello all

I am here to ask for a quick way to find results of numbers with (large) powers.

In one GMAT prep Q

2^5+2^5+3^5+3^5+3^5 can be solved as

2^5(1+1) + 3^5 (1+1+1)
2^6+3^6

But what if we have larger powers
e.g.
3^180 - 3^30 or
3^180 - 3^30 + 2 ^120 - 2^60

What is an easy way to solve such problems with large powers when base has addition and sub.

thanks!

Even the base is small such 2 or 3, such large of powers makes it impossible to write the result. So there can be 2 kinds of questions. First asks the modular correspondance of the number, e.g. what is the last digit of 3^60. Second, it may ask the result with relative to any other number. Let me show it by example.
3^180-3^30 + 2^120-2^60
let x = 3^30 and y = 2^60
then the formula becomes lik
x^6-x + y^3-y
that is x.(x^5-1) +y.(y^2-1)
x(x+1).(x^4-x^3-X^2-x-1) +y(y+1)(y-1)
But you can not do a further thing. Because 2^20 is over 1 million and 2^60 is over one qentrillion(number with 18 digits).

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Manager
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Posts: 65

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Re: Easier way to find powers [#permalink]

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21 Nov 2011, 02:12
maliyeci wrote:
melguy wrote:
Hello all

I am here to ask for a quick way to find results of numbers with (large) powers.

In one GMAT prep Q

2^5+2^5+3^5+3^5+3^5 can be solved as

2^5(1+1) + 3^5 (1+1+1)
2^6+3^6

But what if we have larger powers
e.g.
3^180 - 3^30 or
3^180 - 3^30 + 2 ^120 - 2^60

What is an easy way to solve such problems with large powers when base has addition and sub.

thanks!

Even the base is small such 2 or 3, such large of powers makes it impossible to write the result. So there can be 2 kinds of questions. First asks the modular correspondance of the number, e.g. what is the last digit of 3^60. Second, it may ask the result with relative to any other number. Let me show it by example.
3^180-3^30 + 2^120-2^60
let x = 3^30 and y = 2^60
then the formula becomes lik
x^6-x + y^3-y
that is x.(x^5-1) +y.(y^2-1)
x(x+1).(x^4-x^3-X^2-x-1) +y(y+1)(y-1)
But you can not do a further thing. Because 2^20 is over 1 million and 2^60 is over one qentrillion(number with 18 digits).

Thanks.

I am pretty sure i saw a question somewhere

3^180 - 3^30 and the options i remember were something like

a. 3^180
b. 3^150
c. 3^30
cant remember the other 2.

I had no clue on how to solve such a big power.

Kudos [?]: 16 [0], given: 10

Senior Manager
Status: D-Day is on February 10th. and I am not stressed
Affiliations: American Management association, American Association of financial accountants
Joined: 12 Apr 2011
Posts: 252

Kudos [?]: 370 [0], given: 52

Location: Kuwait
Schools: Columbia university
Re: Easier way to find powers [#permalink]

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21 Nov 2011, 02:12
in this type of problems where you have large exponants, you need to get the comman factor(in another words, factor exponantial terms that have the same base) out and this is how you could simplify it. you factor as how you can do it with equations.
for the 1st example:

3^180 - 3^30=
3^30(3^150-1)

for the second example:

3^180-3^30 + 2^120-2^60

the common factor for the first part is 3^30

3^30(3^150-1)

and for the second part, the common factor is 2^60

2^60(2^60-1)

3^30(3^150-1)+2^60(2^60-1)

_________________

Sky is the limit

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Re: Easier way to find powers [#permalink]

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21 Nov 2011, 02:21
manalq8 wrote:
in this type of problems where you have large exponants, you need to get the comman factor(in another words, factor exponantial terms that have the same base) out and this is how you could simplify it. you factor as how you can do it with equations.
for the 1st example:

3^180 - 3^30=
3^30(3^150-1)

for the second example:

3^180-3^30 + 2^120-2^60

the common factor for the first part is 3^30

3^30(3^150-1)

and for the second part, the common factor is 2^60

2^60(2^60-1)

3^30(3^150-1)+2^60(2^60-1)

thanks for clarifying. I too was thinking along these lines.

Lets take the easier Q .
3^180 - 3^30=
3^30(3^150-1)

How can we solve this thing to get a clean answer. I remember 200% the answer choices were in powers (something like below and not big huge numbers) .

a. 3^180
b 3^150
c 3^30

If we calculate this -> 3^30(3^150-1) .. i think there is no easy solution to the second part - (3^150-1). this is the puzzling part to me. How do we quickly solve this thing - 3^30(3^150-1).

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Re: Easier way to find powers [#permalink]

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21 Nov 2011, 04:23
melguy wrote:
[
Lets take the easier Q .
3^180 - 3^30=
3^30(3^150-1)

How can we solve this thing to get a clean answer. I remember 200% the answer choices were in powers (something like below and not big huge numbers) .

a. 3^180
b 3^150
c 3^30

There is no 'simple' way to write 3^180 - 3^30 as a single power. All you can do is factor it; you can first factor out 3^30, then if you want to go further you can use a difference of squares:

3^180 - 3^30 = 3^30 (3^150 - 1) = 3^30 (3^75 + 1)(3^75 - 1)

To factor that any further, you'd need to know some factorization patterns that aren't tested on the GMAT. Notice that by factoring, we aren't making the answer appear any simpler - in fact, it begins to look more complicated.

So if you're '200% sure' that you saw a similar question in which the answer choices were neat powers for a similar question, you'd need to show us the exact question. The only similar question I could imagine where the answers would be simple powers would be something like:

3^180 - 3^30 is closest to which of the following:

A) 3^180
B) 3^177
C) 3^150
D) 3^30
E) 3^6

This question is not asking for a simplification at all; it's asking for an estimate. Here you simply need to notice that 3^180 is absolutely enormous in comparison to 3^30, so subtracting 3^30 will barely make any difference at all, and the answer is A.
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Re: Easier way to find powers [#permalink]

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21 Nov 2011, 17:30
IanStewart wrote:
melguy wrote:
[
Lets take the easier Q .
3^180 - 3^30=
3^30(3^150-1)

How can we solve this thing to get a clean answer. I remember 200% the answer choices were in powers (something like below and not big huge numbers) .

a. 3^180
b 3^150
c 3^30

There is no 'simple' way to write 3^180 - 3^30 as a single power. All you can do is factor it; you can first factor out 3^30, then if you want to go further you can use a difference of squares:

3^180 - 3^30 = 3^30 (3^150 - 1) = 3^30 (3^75 + 1)(3^75 - 1)

To factor that any further, you'd need to know some factorization patterns that aren't tested on the GMAT. Notice that by factoring, we aren't making the answer appear any simpler - in fact, it begins to look more complicated.

So if you're '200% sure' that you saw a similar question in which the answer choices were neat powers for a similar question, you'd need to show us the exact question. The only similar question I could imagine where the answers would be simple powers would be something like:

3^180 - 3^30 is closest to which of the following:

A) 3^180
B) 3^177
C) 3^150
D) 3^30
E) 3^6

This question is not asking for a simplification at all; it's asking for an estimate. Here you simply need to notice that 3^180 is absolutely enormous in comparison to 3^30, so subtracting 3^30 will barely make any difference at all, and the answer is A.

Thanks a lot for making it clear!

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Re: Easier way to find powers   [#permalink] 21 Nov 2011, 17:30
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