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If \(a\), \(b\), and \(c\) are consecutive integers, where \(10 < a < b < c\), which of the following could be the remainder when \(a^c\) is divided by \(b\)?
I. 1
II. \(b - 1\)
III. \(b\)
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
I. 1
II. \(b - 1\)
III. \(b\)
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
12 Jan 2026, 12:10
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Official Solution:
If \(a\), \(b\), and \(c\) are consecutive integers, where \(10 < a < b < c\), which of the following could be the remainder when \(a^c\) is divided by \(b\)?
I. 1
II. \(b - 1\)
III. \(b\)
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
We have three consecutive integers \(a\), \(b\), and \(c\) with \(10 < a < b < c\). So we can write \(a = b - 1\) and \(c = b + 1\).
We want the remainder when \((b - 1)^{(b + 1)}\) is divided by \(b\).
When you expand this expression, all terms except the last one will be multiples of \(b\) and thus will be divisible by \(b\). The last one will be \((-1)^{(b + 1)}\). Hence, the question essentially boils down to finding the remainder when \((-1)^{(b + 1)}\) is divided by \(b\).
If \(b\) is odd, then \(b + 1\) is even, so \((-1)^{(b + 1)} = 1\). And \(1\) divided by \(b\) will give the remainder of \(1\).
If \(b\) is even, then \(b + 1\) is odd, so \((-1)^{(b + 1)} = -1\). And \(-1\) divided by \(b\) will give the remainder of \(b - 1\). For example, \(-1\) divided by \(20\) gives the remainder of \(20 - 1 = 19\).
Answer: D.
P.S. The process for finding the remainder when dividing a negative integer by a positive integer follows the same principles as when dividing a positive integer by a positive integer.
For example:
• What is the remainder when dividing \(21\) by \(6\)? We find the closest multiple of \(6\) that is less than \(21\), which is \(18\). Then, we calculate \(21 - 18\), yielding a remainder of \(3\).
• What is the remainder when dividing \(-23\) by \(7\)? Here, we find the closest multiple of \(7\) that is less than \(-23\), which is \(-28\). Then, we calculate \(-23 - (-28)\), resulting in a remainder of \(5\).
What about dividing \(-1\) by \(26\)? The closest multiple of \(26\) less than \(-1\) is \(-26\). So, the remainder is \(-1 - (-26) = 25\).
Alternatively, consider this method:
• What is the remainder when dividing \(-23\) by \(7\)? Dividing \(23\) by \(7\) gives a remainder of \(2\). To find the remainder for \(-23\) divided by \(7\), subtract this \(2\) from the divisor. Thus, the remainder when dividing \(-23\) by \(7\) is \(7 - 2 = 5\).
Similarly, what is the remainder when dividing \(-1\) by \(26\)? Dividing \(1\) by \(26\) gives a remainder of \(1\). Therefore, the remainder when dividing \(-1\) by \(26\) is \(26 - 1 = 25\).
Answer: D
If \(a\), \(b\), and \(c\) are consecutive integers, where \(10 < a < b < c\), which of the following could be the remainder when \(a^c\) is divided by \(b\)?
I. 1
II. \(b - 1\)
III. \(b\)
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
We have three consecutive integers \(a\), \(b\), and \(c\) with \(10 < a < b < c\). So we can write \(a = b - 1\) and \(c = b + 1\).
We want the remainder when \((b - 1)^{(b + 1)}\) is divided by \(b\).
When you expand this expression, all terms except the last one will be multiples of \(b\) and thus will be divisible by \(b\). The last one will be \((-1)^{(b + 1)}\). Hence, the question essentially boils down to finding the remainder when \((-1)^{(b + 1)}\) is divided by \(b\).
If \(b\) is odd, then \(b + 1\) is even, so \((-1)^{(b + 1)} = 1\). And \(1\) divided by \(b\) will give the remainder of \(1\).
If \(b\) is even, then \(b + 1\) is odd, so \((-1)^{(b + 1)} = -1\). And \(-1\) divided by \(b\) will give the remainder of \(b - 1\). For example, \(-1\) divided by \(20\) gives the remainder of \(20 - 1 = 19\).
Answer: D.
P.S. The process for finding the remainder when dividing a negative integer by a positive integer follows the same principles as when dividing a positive integer by a positive integer.
For example:
• What is the remainder when dividing \(21\) by \(6\)? We find the closest multiple of \(6\) that is less than \(21\), which is \(18\). Then, we calculate \(21 - 18\), yielding a remainder of \(3\).
• What is the remainder when dividing \(-23\) by \(7\)? Here, we find the closest multiple of \(7\) that is less than \(-23\), which is \(-28\). Then, we calculate \(-23 - (-28)\), resulting in a remainder of \(5\).
What about dividing \(-1\) by \(26\)? The closest multiple of \(26\) less than \(-1\) is \(-26\). So, the remainder is \(-1 - (-26) = 25\).
Alternatively, consider this method:
• What is the remainder when dividing \(-23\) by \(7\)? Dividing \(23\) by \(7\) gives a remainder of \(2\). To find the remainder for \(-23\) divided by \(7\), subtract this \(2\) from the divisor. Thus, the remainder when dividing \(-23\) by \(7\) is \(7 - 2 = 5\).
Similarly, what is the remainder when dividing \(-1\) by \(26\)? Dividing \(1\) by \(26\) gives a remainder of \(1\). Therefore, the remainder when dividing \(-1\) by \(26\) is \(26 - 1 = 25\).
Answer: D
