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# (8^16)+(16^13)+(4^24) = ?

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Intern
Joined: 24 Apr 2012
Posts: 49
Concentration: Strategy
WE: Project Management (Consulting)

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28 Mar 2014, 20:46
6
00:00

Difficulty:

45% (medium)

Question Stats:

67% (01:58) correct 33% (01:46) wrong based on 178 sessions

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(8^16)+(16^13)+(4^24) = ?

A. (4)*(2^29+1)
B. (6)*(2^48)
C. (9)*(2^49)
D. (28)*(2^53)
E. 2^148
VP
Joined: 02 Jul 2012
Posts: 1091
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)

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28 Mar 2014, 21:08
3
(8^16)+(16^13)+(4^24)

((2^3)^16)+((2^4)^13)+((2^2)^24)

(2^48)+(2^52)+(2^48)

(2^48)+((2^48)*(2^4))+(2^48)

(2^48)(1+1+2^4)

(2^48)(2+2^4)

(2^48)*2*(1+2^3)

(2^49)(9)

Intern
Joined: 24 Apr 2012
Posts: 49
Concentration: Strategy
WE: Project Management (Consulting)

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28 Mar 2014, 21:21
Dear MacFauz

I do understand that C is the correct choice but can you please explain why e is wrong?
Intern
Joined: 01 Jan 2013
Posts: 47
Location: India

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28 Mar 2014, 21:26
1
royQV wrote:
Dear MacFauz

I do understand that C is the correct choice but can you please explain why e is wrong?[/quote]

(a^m)(a^n) = a^(m+n)

Eg. 2^3.2^2 = 2^5

However
a^m + a^n is not a^(m+n)
2^3 + 2^2 is not 2^5 (i.e 8 + 4 is not 32)
Intern
Joined: 03 Aug 2012
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30 Mar 2014, 20:53
royQV wrote:
Dear MacFauz

I do understand that C is the correct choice but can you please explain why e is wrong?

I think you are assuming that (X^Y) + (X^Z) is = (X^Y+Z) which is not true, as it only applies to multiplication. Rather it would have to be (X^Y)(X^Z) = (X^YZ)

So when you simplify down to 2^48 + 2^52 + 2^48 you cannot just add exponents.

Here's how my brain works with this one,

Step 1: Recognize a common base.

(8^16) + (16^13) + (4^24) = ((2^2)^16) + ((2^4)^13) + ((2^2)^24) = (2^48) + (2^52) + (2^48)

Step 2: Recognize the factor and pull out of the equation.

= (2^48)(1 + (2^4) + 1)
= (2^48)(1 + 16 + 1)
= (2^48)(18)

(2^48)(18) = (2^48)(2)(9) = (2^49)(9)

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31 Mar 2014, 18:11
2
1
$$8^{16} = 2^{48}$$

$$16^{13} = 2^{52}$$

$$4^{24} = 2^{48}$$

$$2^{48} + 2^{52} + 2^{48}$$

$$= 2^{49} + 2^{52}$$

$$= 2^{49} (1+8)$$

$$= 9 * 2^{49}$$

Director
Joined: 23 Jan 2013
Posts: 507
Schools: Cambridge'16

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31 Mar 2014, 20:18
royQV wrote:
Dear MacFauz

I do understand that C is the correct choice but can you please explain why e is wrong?[/quote]

You cannot just add exponents when you sum the same bases, in this case 2. If it was multiplying you could do so. Another rule which is useful here is that 2^48+2^48=2^49. Adding as much times as base (the same) and same exponent gives us the same base with +1 in exponent
Math Expert
Joined: 02 Sep 2009
Posts: 64151

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01 Apr 2014, 01:12
royQV wrote:

I do understand that C is the correct choice but can you please explain why e is wrong?

Theory on Exponents: math-number-theory-88376.html

All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39
All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60

Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html

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Joined: 17 Nov 2013
Posts: 75

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24 Nov 2016, 09:35
In this question, the trick is to do the below

2^48 + 2^48 = 2^49

I did not know the above was possible before. I will remember this for future questions.
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06 Apr 2020, 17:37
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Re: (8^16)+(16^13)+(4^24) = ?   [#permalink] 06 Apr 2020, 17:37