If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 =

Question

If x and y are nonnegative integers and 2,008^{2,008} - 2,008^{2,007} = (2,008^x)(2,007^y) What is the value of x + y? (A) 1 (B) 2 (C) 2,007 (D) 2,008 (E) 4,015 Captura.PNG

gmatophobia · Accepted Answer

2008^{2008} - 2008^{2007} = (2008^x)*(2007^y)

2008^{2007}(2008 - 1) = (2008^x)*(2007^y)

2008^{2007}(2007) = (2008^x)*(2007^y)

Hence, x = 2007 ; y = 1

x + y = 2007 + 1 = 2008

Option D

smseba · Answer

Can someone please explain how to solve this? I could not understand the explanation provided.

Krunaal · Answer

2008^{2008} - 2008^{2007} = (2008^x)*(2007^y)

We can write  2008^{2008}  as  2008^{2007+1}

2008^{2007+1} - 2008^{2007} = (2008^x)*(2007^y)

We can write  2008^{2007+1}  as  2008^{2007} * 2008   ................. since  a^{m+n} = a^{m} * a^{n}

2008^{2007} * 2008 - 2008^{2007} = (2008^x)*(2007^y)

Taking  2008^{2007}  common

2008^{2007}*(2008 - 1) = (2008^x)*(2007^y)

2008^{2007}*(2007) = (2008^x)*(2007^y)

Comparing LHS and RHS

Hence, x = 2007 ; y = 1

x + y = 2007 + 1 = 2008

Hope it helps. Let me know if you did not follow any specific step.

sujoykrdatta · Answer

2,008^2,008 - 2,008^2,007 = 2008^2007 * 2008^1 - 2008^2007 
= 2008^2007 * ( 2008^1 - 1) 
= 2008^2007 * 2007

=> 2008^x * 2007^y = 2008^2007 * 2007^1 
=> x = 2007 and y = 1 
=> x + y = 2008

Ans D

DanTheGMATMan · Answer

Classic exponents problem. Easy to do:
 https://www.youtube.com/watch?v=BxcbXlxTG30

lawyeroffduty · Answer

can anyone please share some more similar type of questions, and this is the question from the official mock 3

Bunuel · Answer

Exponents and Roots of Numbers

Theory: 
 
  Exponents and Roots on the GMAT: Tips and hints  
  Cyclicity and the remainders on the GMAT

Questions 
 
  Units digits, exponents, remainders problems  
  Tough and tricky DS exponents and roots questions with detailed solutions  
  Tough and tricky PS exponents and roots questions with detailed solutions  
  Exponents PS Questions  
  Roots PS Questions

Check below for more: 
 
  ALL YOU NEED FOR QUANT ! ! !  
  Ultimate GMAT Quantitative Megathread

Hope this helps.

georgec · Answer

To make this simpler, substitute 2,008 as a. We get:
a^a - a^(a-1) = a^x * (a-1)^y;
a^(a-1) * (a^1 - 1) = a^x * (a-1)^y;
a^(a-1) * (a-1)^1 = a^x * (a-1)^y;
Do you see it now?
a-1=x and y=1.
Then x=2,008-1=2,007 and y=1. The sum is 2,007+1=2,008.

RobertVericPrep · Answer

When you see questions that involve exponents, you should always think about factoring. That is perhaps the biggest thing being tested in this question.

The equation is:

2008^2008 - 2008^2007 = (2008^x)(2007^y)

On the left-hand side of the equation, you can factor out 2008^2007 from both terms because it is a common factor. (Just as 2^2 is a common factor of 2^5 and 2^3, for example.)

This gives us:

2008^2007 * (2008 - 1)

which equals:

2008^2007 * (2007)

which is the same as:

(2008^2007) * (2007^1)

Now let’s compare that with the right-hand side of the equation, which was

(2008^x) * (2007^y)

You can see that x corresponds with the exponent 2007, and y corresponds with the exponent 1.

x = 2007
y = 1

Therefore:
x + y = 2007 + 1 = 2008

Answer: D

If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 =

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GMAT Problem Solving (PS) Questions

If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 =

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards