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01 Feb 2024, 02:04
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B
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Difficulty:
75%
(hard)
Question Stats:
62% (02:35) correct
38%
(02:42)
wrong
based on 204
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If the sum of all \(x^n\), where x is a positive two-digit integer and n is an integer from 0 to 5 inclusive, equals k, which of the following equals x^6?
A. kx + 1
B. k(x - 1) + 1
C. k(x - 1) - 1
D. k(x - 2) + 1
E. k(x - 2) - 1
A. kx + 1
B. k(x - 1) + 1
C. k(x - 1) - 1
D. k(x - 2) + 1
E. k(x - 2) - 1
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27 Feb 2024, 02:53
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Official Solution:
If the sum of all \(x^n\), where \(x\) is a positive two-digit integer and \(n\) is an integer from 0 to 5 inclusive, equals \(k\), which of the following equals \(x^6\)?
A. \(kx + 1\)
B. \(k(x - 1) + 1\)
C. \(k(x - 1) - 1\)
D. \(k(x - 2) + 1\)
E. \(k(x - 2) - 1\)
We are given that:
This simplifies to:
Multiply both sides of the equation by \(x\):
Adjust the equation by adding and subtracting 1 from the left-hand side:
Substitute \(1 + x + x^2 + x^3 + x^4 + x^5 = k\) back into our equation:
Answer: B
If the sum of all \(x^n\), where \(x\) is a positive two-digit integer and \(n\) is an integer from 0 to 5 inclusive, equals \(k\), which of the following equals \(x^6\)?
A. \(kx + 1\)
B. \(k(x - 1) + 1\)
C. \(k(x - 1) - 1\)
D. \(k(x - 2) + 1\)
E. \(k(x - 2) - 1\)
We are given that:
- \(x^0 + x^1 + x^2 + x^3 + x^4 + x^5 = k\)
This simplifies to:
- \(1 + x + x^2 + x^3 + x^4 + x^5 = k\)
Multiply both sides of the equation by \(x\):
- \(x + x^2 + x^3 + x^4 + x^5 + x^6 = kx\)
Adjust the equation by adding and subtracting 1 from the left-hand side:
- \(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 - 1 = kx\)
Substitute \(1 + x + x^2 + x^3 + x^4 + x^5 = k\) back into our equation:
- \(k + x^6 - 1 = kx\)
\(x^6 = kx - k + 1\)
\(x^6 = k(x - 1) + 1\)
Answer: B
General Discussion
01 Feb 2024, 02:07
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Try solving by straightforward algebraic manipulation, avoiding overthinking and overcomplicating.
