5(2^4)(3^32) =

Question

5(2^4)(3^{32}) = (A) 5(6^{36}) (B) (10^4)(3^{32}) (C) (3^{34}) + (3^{32}) (D) (3^{35}) - (3^{33}) (E) (3^{36}) - (3^{32}) For a set of challenging GMAT problems on exponents and roots, as well as the OE for this problem, see: https://magoosh.com/gmat/2014/challengin ... and-roots/ Mike :-)

Celestial09 · Accepted Answer

IMO its E
firstly in now way we could use or modify 3^32
secondly a trick of using negative remainder. 
so i used 2^4 with 5.. though after spending 1 min on it that its 80 and hence divisible by 3^4 as negative remainder of -1 
so its 81-1... which is needed to be used
now here i got confused between D AND E
i did eliminated D simply because of its odd powers which didn't gave much sense to me as its a 2mins time constraint.
so hence i arrived at E
81 is 3^4 so no use of odd powers else result would be negative elsewise.

this is my first time i solved a question and gave a bit explanation too.

kudos will entice me to work hard...

PareshGmat · Answer

Answer = (E)  (3^{36}) - (3^{32})

5 * 2^4 * 3^{32}

= 80 * 3^{32}

= (81 - 1) * 3^{32}

= 3^{36} - 3^{32}

Vaibhav21 · Answer

E.
Ques - 5 X (2^4=16) X 3^32 = 80 X 3^32
E = 3^32 x (3^4 - 1) = 80 x 3^32

Celestial09 · Answer

Hello!
Another way out

5(2^4)(s^32)= 80(81^8)
as 3^4 is 81

80 can be written as 81-1
(81-1)(81^8)
(81^9)-81^8
3^36-3^32

Hence E
this is the simplest i could figured out

kindly share that binomial eq in imp for gmat?

pacifist85 · Answer

Yep I also did it similarly.

Starting with the question stem:
5 * 2^4 * 3^32
5 * 2^1 * 2^3 * 3^32
10 * 2^3 * 3^32
80 * 3^32

Then moving on to the answer choices, A and B are relatively easy to eliminate. Plus, the fact that we see additions and subtractions should alert as to the possibility of taking a common factor and creating a multiplication as an effect.

I started witd D which resulted in 3^33 * 8. Close, but not what I was looking for. So I moved to E, which results in:
3^36 - 3^32
3^32 (3^4 - 1)
3^32 (81 - 1)
3^32 * 80, which is what we are looking for. So,  ANS E

Beat720 · Answer

5 * 2^4 * 3^{32}

= 10*8 * 3^{32}

= (3^2 + 1)*(3^2-1) * 3^{32}

= (3^4 - 1) * 3^{32}

= 3^{36} - 3^{32}

--> Answer E

KanishkM · Answer

Quite an intriguing one.

5(2^4)(3^{32})  = 80 *  (3^{32})

Now just browsing over the options you can get the value as E.
(A)  5(6^{36}) ------------------------------------> 5* 2^{36} *3^36
(B)  (10^4)(3^{32}) -----------------------------> wont give 80
(C)  (3^{34}) + (3^{32}) -----------------------> wont give 80
(D)  (3^{35}) - (3^{33}) ------------------------> wont give 80
(E)  (3^{36}) - (3^{32}) ------------------------> took  3^{32}  common, giving  3^4 - 1 =80

Answer E.

bumpbot · Answer

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5(2^4)(3^32) =

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