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Sub 505 (Easy)|   Arithmetic|   Exponents|                     
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Bunuel
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Bunuel


17^3 + 17^4 =

(A) 17^7
(B) 17^3*18
(C) 17^6*18
(D) 2*17^3 + 17
(E) 2*17^3 - 17
\(17^3 + 17^4\)

Or, \(17^3( 1 + 17)\)

Or, \(17^3*18\), thus answer will be (B)
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WRT Question
17^3+17^4
= 17^3(common) [1+17]
=17^3*18
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Bunuel
17^3 + 17^4 =

(A) 17^7
(B) 17^3*18
(C) 17^6*18
(D) 2*17^3 + 17
(E) 2*17^3 - 17

17^3 + 17^4 = 17^3(1 + 17) = 17^3 x 18

Answer: B
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Bunuel
SOLUTION

17^3 + 17^4 =

(A) 17^7
(B) 17^3*18
(C) 17^6*18
(D) 2*17^3 + 17
(E) 2*17^3 - 17

\(17^3 + 17^4 =17^3(1+17)=17^3*18\).

Answer: B.

Hi Bunuel, how do you call this rule when from this \(17^3 + 17^4\) you get this \(17^3(1+17)\) ? please explain in two or three words :)
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dave13
Bunuel
SOLUTION

17^3 + 17^4 =

(A) 17^7
(B) 17^3*18
(C) 17^6*18
(D) 2*17^3 + 17
(E) 2*17^3 - 17

\(17^3 + 17^4 =17^3(1+17)=17^3*18\).

Answer: B.

Hi Bunuel, how do you call this rule when from this \(17^3 + 17^4\) you get this \(17^3(1+17)\) ? please explain in two or three words :)

We are factoring out common term 17^3 from 17^3 + 17*17^3 to get 17^3(1+17).
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Bunuel
17^3 + 17^4 =

(A) 17^7
(B) 17^3*18
(C) 17^6*18
(D) 2*17^3 + 17
(E) 2*17^3 - 17

17^3 + 17^4 = 17^3 (1+ 17) = 17^3 (18)

Answer: B
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Bunuel
17^3 + 17^4 =

(A) 17^7
(B) 17^3*18
(C) 17^6*18
(D) 2*17^3 + 17
(E) 2*17^3 - 17
17^3 + 17^4

Or, 17^3(1 + 17)

17^3 * 18

(B)
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Bunuel
The Official Guide For GMAT® Quantitative Review, 2ND Edition

\(17^3 + 17^4\) =

(A) 17^7
(B) 17^3*18
(C) 17^6*18
(D) 2*17^3 + 17
(E) 2*17^3 - 17

\(17^3 + 17^4\) has the same format as \(x^3 + x^4\), which we can factor as follows: \(x^3(1 + x)\)
The same applies to the given expression.
\(17^3 + 17^4=17^3(1 + 17)\)
\(=(17^3)(18)\)

Answer: B

Cheers,
Brent
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Can someone please explain why the answer still cannot be A?
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meganmcl1
Can someone please explain why the answer still cannot be A?

When you multiply powers with the same base you add the exponents. When you add powers with the same base you can only take common factor.
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andreagonzalez2k
Thank you so much!
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\(17^3 + 17^4 = 17^3(1 + 17) = 17^3(18)\)

Answer B
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Bunuel
17^3 + 17^4 =

(A) 17^7
(B) 17^3*18
(C) 17^6*18
(D) 2*17^3 + 17
(E) 2*17^3 - 17

Hello, people. For this question, making everything in the expression have a power of 3 can be helpful.

\(17^3 + 17^4\)

\(17^3 + 17(17^3)\)

\((1)(17^3) + (17)(17^3)\)

\((18)(17^3)\)

(B) is your answer.
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Bunuel
17^3 + 17^4 =

(A) 17^7
(B) 17^3*18
(C) 17^6*18
(D) 2*17^3 + 17
(E) 2*17^3 - 17





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