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16 Sep 2014, 01:10
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If \(x\) and \(y\) are positive integers, is \(\frac{10^y-x}{3}\) an integer?
(1) \(\frac{x-1}{3}\) is an integer.
(2) \(\frac{y+1}{3}\) is an integer
(1) \(\frac{x-1}{3}\) is an integer.
(2) \(\frac{y+1}{3}\) is an integer
16 Sep 2014, 01:10
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Official Solution:
If \(x\) and \(y\) are positive integers, is \(\frac{10^y-x}{3}\) an integer?
For \(\frac{10^y-x}{3}\) to be an integer, \(10^y-x\) must be divisible by 3. This means the sum of the digits of \(10^y-x\) must also be divisible by 3. Since the sum of the digits of \(10^y\) is always 1 (for positive integer \(y\)), the sum of the digits of \(10^y-x\) will be divisible by 3 if \(x\) is: 1, 4, 7, 10, ... and so on. Essentially, \(x\) needs to be one more than a multiple of 3.
(1) \(\frac{x-1}{3}\) is an integer.
Given: \(\frac{x-1}{3}\) is an integer. This implies \(x = 3*\text{integer} + 1\), meaning \(x\) is one more than a multiple of 3. Sufficient.
(2) \(\frac{y+1}{3}\) is an integer.
There's no information about \(x\), so this is insufficient.
Answer: A
If \(x\) and \(y\) are positive integers, is \(\frac{10^y-x}{3}\) an integer?
For \(\frac{10^y-x}{3}\) to be an integer, \(10^y-x\) must be divisible by 3. This means the sum of the digits of \(10^y-x\) must also be divisible by 3. Since the sum of the digits of \(10^y\) is always 1 (for positive integer \(y\)), the sum of the digits of \(10^y-x\) will be divisible by 3 if \(x\) is: 1, 4, 7, 10, ... and so on. Essentially, \(x\) needs to be one more than a multiple of 3.
(1) \(\frac{x-1}{3}\) is an integer.
Given: \(\frac{x-1}{3}\) is an integer. This implies \(x = 3*\text{integer} + 1\), meaning \(x\) is one more than a multiple of 3. Sufficient.
(2) \(\frac{y+1}{3}\) is an integer.
There's no information about \(x\), so this is insufficient.
Answer: A
11 Oct 2023, 01:22
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
