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505-555 (Easy)|   Exponents|      
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pablovaldesvega
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If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 = Updated on: 24 Jan 2024, 08:46
Originally posted by pablovaldesvega on 19 Nov 2023, 02:33.
Last edited by Bunuel on 24 Jan 2024, 08:46, edited 3 times in total.
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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

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If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 = 19 Nov 2023, 11:54
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pablovaldesvega
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

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

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

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

Hence, x = 2007 ; y = 1

x + y = 2007 + 1 = 2008

Option D
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smseba
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If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 = 11 Jan 2025, 12:19
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Can someone please explain how to solve this? I could not understand the explanation provided.
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If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 = 11 Jan 2025, 12:51
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smseba
Can someone please explain how to solve this? I could not understand the explanation provided.
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\(2008^{2008} - 2008^{2007} = (2008^x)*(2007^y)\)

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.
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If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 = 13 Jan 2025, 07:05
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pablovaldesvega
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
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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
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If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 = 13 Jan 2025, 11:29
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Classic exponents problem. Easy to do:
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lawyeroffduty
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If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 = 27 Jan 2026, 04:53
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can anyone please share some more similar type of questions, and this is the question from the official mock 3
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Bunuel
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If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 = 27 Jan 2026, 07:27
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lawyeroffduty
can anyone please share some more similar type of questions, and this is the question from the official mock 3
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Exponents and Roots of Numbers

Theory:

Questions

Check below for more:

Hope this helps.­
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If x and y are nonnegative integers and 2,008^2,008 - 2,008^2,007 = 25 Mar 2026, 19:02
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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.
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