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Adding exponents problem from pract. Test

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Junior | Next Rank: 30 Posts
Posts: 26
Joined: Fri Apr 06, 2007 10:40 am
by sev » Fri Jun 08, 2007 4:03 pm
2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 =


A )2^9

B) 2^10

C) 2^ 16

D) 2^ 35

E) 2^37

Obviously it's not E, but how do you do this problem? The OA is A which is 2^9

Is there any way to do it without actually calculating all of the values, cuz that just doesnt seem right to me.
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Source: — Problem Solving |
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Tue May 29, 2007 10:52 pm
Location: Pune, India

its not that difficult

by smashingdon » Mon Jun 11, 2007 4:10 am
even if you work it out its simple.
expand just the first few ones

it will become
2 + 2 + 4 + 8 + 16 + 2^5 + 2^6 + 2^7 + 2^8
will become
32 + 2^5 + 2^6 + 2^7 + 2^8
2^5 + 2^5 + 2^6 + 2^7 + 2^8
2^5[1+1+2+4+8]
2^5 * 16
2^5 * 2^4
2^9
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Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Mon Mar 12, 2007 9:14 pm

Re: Adding exponents problem from pract. Test

by bills_e » Tue Jul 17, 2007 2:55 pm
sev wrote:2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 =


A )2^9

B) 2^10

C) 2^ 16

D) 2^ 35

E) 2^37

Obviously it's not E, but how do you do this problem? The OA is A which is 2^9

Is there any way to do it without actually calculating all of the values, cuz that just doesnt seem right to me.
2+2 = 2^2. (Adding the first two terms)
2^2 + 2^2 = 2^3 (Adding the above result to the third term)
2^3 + 2^3 = 2^4 (Adding the above result to the fourth term)

.....

2^8+2^8 = 2^9

Therefore, A.
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