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16 Sep 2014, 01:45
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When a positive integer \(x\) is divided by a positive integer \(y\), the quotient is 3 and the remainder is \(z\). When \(z\) is divided by \(y\), the remainder is 2. Which of the following could be the value of \(x\)?
I. 5
II. 8
III. 32
A. I only
B. II only
C. III only
D. I and II only
E. I, II and III
I. 5
II. 8
III. 32
A. I only
B. II only
C. III only
D. I and II only
E. I, II and III
16 Sep 2014, 01:45
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Official Solution:
When a positive integer \(x\) is divided by a positive integer \(y\), the quotient is 3 and the remainder is \(z\). When \(z\) is divided by \(y\), the remainder is 2. Which of the following could be the value of \(x\)?
I. 5
II. 8
III. 32
A. I only
B. II only
C. III only
D. I and II only
E. I, II and III
We are given that a positive integer \(x\) divided by a positive integer \(y\) gives a quotient of 3 and a remainder of \(z\). This can be expressed as \(x = 3y + z\), where \(0 \leq z < y\) (since the remainder is always less than the divisor).
The problem also states that when \(z\) is divided by \(y\), the remainder is 2. Since the remainder is always less than the divisor, it follows that \(2 < y\). Furthermore, we know that when the divisor (\(y\) in this case) is greater than the dividend (\(z\) in this case), the remainder must equal the dividend. Therefore, we can deduce that \(z = 2\).
Substituting \(z = 2\) in \(x = 3y + z\), we get \(x = 3y + 2\). As \(y\) is a positive integer and \(2 < y\), the smallest possible value for \(y\) is 3. Thus, the minimum value of \(x\) is \(x = 3*3 + 2 = 11\). Therefore, \(x\) cannot be 5 or 8.
To check if \(x\) could be 32, we can set \(y = 10\), which satisfies the condition that \(2 < y\). Substituting \(y = 10\) in \(x = 3y + 2\), we get \(x = 3*10 + 2 = 32\). Therefore, \(x\) could be 32, and option III is a possible value for \(x\).
Answer: C
When a positive integer \(x\) is divided by a positive integer \(y\), the quotient is 3 and the remainder is \(z\). When \(z\) is divided by \(y\), the remainder is 2. Which of the following could be the value of \(x\)?
I. 5
II. 8
III. 32
A. I only
B. II only
C. III only
D. I and II only
E. I, II and III
We are given that a positive integer \(x\) divided by a positive integer \(y\) gives a quotient of 3 and a remainder of \(z\). This can be expressed as \(x = 3y + z\), where \(0 \leq z < y\) (since the remainder is always less than the divisor).
The problem also states that when \(z\) is divided by \(y\), the remainder is 2. Since the remainder is always less than the divisor, it follows that \(2 < y\). Furthermore, we know that when the divisor (\(y\) in this case) is greater than the dividend (\(z\) in this case), the remainder must equal the dividend. Therefore, we can deduce that \(z = 2\).
Substituting \(z = 2\) in \(x = 3y + z\), we get \(x = 3y + 2\). As \(y\) is a positive integer and \(2 < y\), the smallest possible value for \(y\) is 3. Thus, the minimum value of \(x\) is \(x = 3*3 + 2 = 11\). Therefore, \(x\) cannot be 5 or 8.
To check if \(x\) could be 32, we can set \(y = 10\), which satisfies the condition that \(2 < y\). Substituting \(y = 10\) in \(x = 3y + 2\), we get \(x = 3*10 + 2 = 32\). Therefore, \(x\) could be 32, and option III is a possible value for \(x\).
Answer: C
General Discussion
