A repunit is a positive integer that contains only the digit 1. If the integer r is a repunit, is r prime?
(1) The number of digits in r is a multiple of 3.
(2) 0 < r < 1200
What's the best way to determine whether statement 1 is sufficient?
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A repunit is a positive integer that contains only the digit
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Statement 1 tells us that the number may be 111 or 111,111 or even 111,111,111.
If you know the divisibility rules for 3, then you will know that each of these numbers is divisible by 3.
To determine whether a number is divisible by 3, add up the unit digits. If the result is divisible by 3, then the number is divisible by 3.
1+1+1 = 3, so the number 111 is divisible by 3.
1+1+1+1+1+1 = 6, which is divisible by 3. Accordingly, the number 111,111 is also divisible by 3.
So statement 1 tells us that the number is evenly divisible by 3, and thus not a prime number. Accordingly, statement 1 is sufficient on its own.
If you know the divisibility rules for 3, then you will know that each of these numbers is divisible by 3.
To determine whether a number is divisible by 3, add up the unit digits. If the result is divisible by 3, then the number is divisible by 3.
1+1+1 = 3, so the number 111 is divisible by 3.
1+1+1+1+1+1 = 6, which is divisible by 3. Accordingly, the number 111,111 is also divisible by 3.
So statement 1 tells us that the number is evenly divisible by 3, and thus not a prime number. Accordingly, statement 1 is sufficient on its own.
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Target question: Is r prime?ardz24 wrote:A repunit is a positive integer that contains only the digit 1. If the integer r is a repunit, is r prime?
(1) The number of digits in r is a multiple of 3.
(2) 0 < r < 1200
Given: Integer r is a repunit
So, some possible values of r include: 1, 11, 111, 1111, 111111111111111111, etc
Statement 1: The number of digits in r is a multiple of 3.
So, r could equal 111 or 111,111 or 111,111,111 or 111,111,111,111 etc
We know that, if the sum of the digits in a number is divisible by 3, then that number is also divisible by 3
For example, since the digits in 111 add to 3 (and 3 is divisible by 3), we know that 111 is divisible by 3, which means 111 is NOT prime.
Likewise, since the digits in 111,111 add to 6 (and 6 is divisible by 3), we know that 111,111 is divisible by 3, which means 111,111 is NOT prime.
And, since the digits in 111,111,111 add to 9 (and 9 is divisible by 3), we know that 111,111,111 is divisible by 3, which means 111,111,111 is NOT prime.
As we can see, if the number of digits in r is a multiple of 3, then the sum of r's digits will be divisible by 3.
So, the answer to the target question is NO, r is NOT prime
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: 0 < r < 1200
There are 4 possible values of r: 1 or 11 or 111 or 1111
Case a: If r = 11, then r IS prime
Case b:If r = 111, then r is NOT prime
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Answer: A
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Brent
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