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19 Aug 2022, 08:13
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If \(k\) is a positive integer, is \(\sqrt{k}\) an integer ?
(1) \(k\) is a multiple of every single-digit prime number.
(2) The tens digit of \(k\) is a factor of a single digit prime number.
(1) \(k\) is a multiple of every single-digit prime number.
(2) The tens digit of \(k\) is a factor of a single digit prime number.
19 Aug 2022, 08:13
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Official Solution:
If \(k\) is a positive integer, is \(\sqrt{k}\) an integer ?
(1) \(k\) is a multiple of every single-digit prime number.
There are four single-digit prime numbers: 2, 3, 5, and 7. So, \(k\) is a multiple of \(2*3*5*7\). This statement is clearly insufficient. For example, if \(k=2*3*5*7\), then the answer is NO, but if \(k=(2*3*5*7)^2\), then the answer is YES. Not sufficient.
Note that, from this statement, we can deduce that the units digit of \(k\) is 0.
(2) The tens digit of \(k\) is a factor of a single digit prime number.
There are four single-digit prime numbers: 2, 3, 5, and 7. So, the tens digit of \(k\) is 1, 2, 3, 5, or 7. This statement is also clearly insufficient. For example, if \(k=20\), then the answer is NO, but if \(k=25\), then the answer is YES. Not sufficient.
(1)+(2) We know that the units digit of \(k\) is 0 (from statement 1) and the tens digit of \(k\) is 1, 2, 3, 5, or 7 (from statement 2).
Since the units digit of \(k\) is 0, it is divisible by 10, which means that \(k\) is divisible by both 2 and 5. For \(\sqrt{k}\) to be an integer, \(k\) must be a perfect square. In order to be a perfect square, the powers of the prime factors in \(k\) must be even. Thus, both 2 and 5 should have even powers (2, 4, 6, ...). Consequently, \(k\) must have at least two factors of both 2 and 5 (i.e., 2^2 and 5^2), and therefore must be divisible by 100. In this case, both the units digit and the tens digit of \(k\) would need to be 0. In other words, for \(k\) to be a perfect square, it must be divisible not only by 10 but also by \(10^2=100\), and thus must have both the units and tens digits equal to 0.
However, from statement 2, we know that the tens digit of \(k\) is NOT 0. Therefore, \(\sqrt{k}\) cannot be an integer and we have a definite NO answer to the question. Sufficient.
Answer: C
If \(k\) is a positive integer, is \(\sqrt{k}\) an integer ?
(1) \(k\) is a multiple of every single-digit prime number.
There are four single-digit prime numbers: 2, 3, 5, and 7. So, \(k\) is a multiple of \(2*3*5*7\). This statement is clearly insufficient. For example, if \(k=2*3*5*7\), then the answer is NO, but if \(k=(2*3*5*7)^2\), then the answer is YES. Not sufficient.
Note that, from this statement, we can deduce that the units digit of \(k\) is 0.
(2) The tens digit of \(k\) is a factor of a single digit prime number.
There are four single-digit prime numbers: 2, 3, 5, and 7. So, the tens digit of \(k\) is 1, 2, 3, 5, or 7. This statement is also clearly insufficient. For example, if \(k=20\), then the answer is NO, but if \(k=25\), then the answer is YES. Not sufficient.
(1)+(2) We know that the units digit of \(k\) is 0 (from statement 1) and the tens digit of \(k\) is 1, 2, 3, 5, or 7 (from statement 2).
Since the units digit of \(k\) is 0, it is divisible by 10, which means that \(k\) is divisible by both 2 and 5. For \(\sqrt{k}\) to be an integer, \(k\) must be a perfect square. In order to be a perfect square, the powers of the prime factors in \(k\) must be even. Thus, both 2 and 5 should have even powers (2, 4, 6, ...). Consequently, \(k\) must have at least two factors of both 2 and 5 (i.e., 2^2 and 5^2), and therefore must be divisible by 100. In this case, both the units digit and the tens digit of \(k\) would need to be 0. In other words, for \(k\) to be a perfect square, it must be divisible not only by 10 but also by \(10^2=100\), and thus must have both the units and tens digits equal to 0.
However, from statement 2, we know that the tens digit of \(k\) is NOT 0. Therefore, \(\sqrt{k}\) cannot be an integer and we have a definite NO answer to the question. Sufficient.
Answer: C
09 Sep 2022, 22:45
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Hi @Banuel,
Didn't understand the following line -
"For √k to be an integer, the tens digit of k must also be 0 (so k must be divisible not only by 10 but also by 100".
What's the concept here? If unit digit is 0, can there be no case where tens digit is not 0 but the number is a square?
Didn't understand the following line -
"For √k to be an integer, the tens digit of k must also be 0 (so k must be divisible not only by 10 but also by 100".
What's the concept here? If unit digit is 0, can there be no case where tens digit is not 0 but the number is a square?
