If n is a positive integer, which of the following can’t b

[Max@Math Revolution]

[Math Revolution GMAT math practice question]

If n is a positive integer, which of the following can't be the value of (n+1)^4-n^4?

A. 2465
B. 4641
C. 6096
D. 7825
E. 9855

[bubbliiiiiiii]

Odd * odd = odd.

Simplify the equation using (a+b)(a-b) = a^2 - b^2.

[Max@Math Revolution]

=>


If n + 1 is an even number, then n is an odd number and (n+1)^4-n^4 must be an odd number.
If n + 1 is an odd number, then n is an even number and (n+1)^4-n^4 must be an odd number.
All answer choices except for C) are odd numbers.

Therefore, the answer is C.

Answer: C

[Scott@TargetTestPrep]

If n is a positive integer, which of the following can't be the value of (n+1)^4-n^4?

A. 2465
B. 4641
C. 6096
D. 7825
E. 9855

Let's simplify the expression as a difference of squares:

(n+1)^4-n^4

[(n + 1)^2 - n^2][(n + 1)^2 + n^2]

[(n + 1) - n][(n + 1) + n][(n + 1)^2 + n^2]

[1][2n + 1][(n + 1)^2 + n^2]

[2n + 1][(n + 1)^2 + n^2]

We see that (2n + 1) is odd regardless of whether n is odd or even. Similarly, (n + 1)^2 + n^2 must be odd since the two terms are squares of consecutive integers and hence one of them must be even and the other must be odd. Since both 2n + 1 and (n + 1)^2 + n^2 are odd, their product is odd. So it can't be 6096.

Alternate solution:

We see that n and n + 1 are consecutive integers. So if n is odd, then n + 1 is even. Since a positive integer raised to a positive integer power has the same parity (i.e., odd or even) as the integer base. (n + 1)^4 is even and n^4 is odd, so their difference is odd. Similarly, if n is even, then n + 1 is odd. Furthermore, (n + 1)^4 is odd and n^4 is even and their difference will still be odd.

Answer: C

[Brent@GMATPrepNow]

[Math Revolution GMAT math practice question]

If n is a positive integer, which of the following can't be the value of (n+1)^4 - n^4?

A. 2465
B. 4641
C. 6096
D. 7825
E. 9855

KEY CONCEPTS:
#1) If n is an integer, then n and n+1 are consecutive integers
#2)If n and n+1 are consecutive integers, then one value is ODD and the other value is EVEN
#3)ODD^4 = (ODD)(ODD)(ODD)(ODD) = ODD
#4)EVEN^4 = (EVEN)(EVEN)(EVEN)(EVEN) = EVEN
#5)(odd)-(odd) = even
#6)(odd)-(even) = odd
#7)(even)-(odd) = odd
#8)(even)-(even) = even

From #2, there are two possible cases to consider:
case 1: n is EVEN and n+1 is ODD
case 2: n is ODD and n+1 is EVEN

case 1: n is EVEN and n+1 is ODD
In this case, (n+1)^4 - n^4 = ODD^4 - EVEN^4
= ODD - EVEN (from #3 and #4)
= ODD (from #6)

case 2: n is ODD and n+1 is EVEN
In this case, (n+1)^4 - n^4 = EVEN^4 - ODD^4
= EVEN - ODD (from #3 and #4)
= ODD (from #7)

In both cases, (n+1)^4 - n^4 = some ODD integer.

Check the answer choices........
All of the answer choices are ODD, except for answer choice C

Since (n+1)^4 - n^4 can't be EVEN, the correct answer is C

Cheers,
Brent