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07 Nov 2023, 09:33
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What is the remainder when \(11^1 + 11^2 + 11^3 + ... + 11^{11}\) is divided by 22 ?
A. 0
B. 1
C. 10
D. 11
E. 21
A. 0
B. 1
C. 10
D. 11
E. 21
07 Nov 2023, 09:33
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Official Solution:
What is the remainder when \(11^1 + 11^2 + 11^3 + ... + 11^{11}\) is divided by 22 ?
A. 0
B. 1
C. 10
D. 11
E. 21
\(11^1 + 11^2 + 11^3 + ... + 11^{11}=\)
\(=11^1 + 11(11 + 11^2 + ... + 11^{10})\)
Observe that within the parentheses, we have the sum of 10 odd numbers, which results in an even number. Thus:
\(11^1 + 11(11 + 11^2 + ... + 11^{10})=\)
\(=11^1 + 11*\text{{even}}=\)
\(=11^1 + \text{{a multiple of 22}}\)
Therefore, the remainder when the above expression is divided by 22 is 11.
Answer: D
What is the remainder when \(11^1 + 11^2 + 11^3 + ... + 11^{11}\) is divided by 22 ?
A. 0
B. 1
C. 10
D. 11
E. 21
\(11^1 + 11^2 + 11^3 + ... + 11^{11}=\)
\(=11^1 + 11(11 + 11^2 + ... + 11^{10})\)
Observe that within the parentheses, we have the sum of 10 odd numbers, which results in an even number. Thus:
\(11^1 + 11(11 + 11^2 + ... + 11^{10})=\)
\(=11^1 + 11*\text{{even}}=\)
\(=11^1 + \text{{a multiple of 22}}\)
Therefore, the remainder when the above expression is divided by 22 is 11.
Answer: D
15 Mar 2024, 14:50
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Can you explain why the answer isn't 1?
My rationale:
1. Factor out 11 from the numerator
2. Divide numerator and denominator by 11
3. Now you have numerator / 2. Since the numerator is odd and an odd # divided by 2 has a remaindor of 1, remaindor=1.
My rationale:
1. Factor out 11 from the numerator
2. Divide numerator and denominator by 11
3. Now you have numerator / 2. Since the numerator is odd and an odd # divided by 2 has a remaindor of 1, remaindor=1.
