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14 Jul 2022, 08:00
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Question Stats:
42% (02:03) correct
58%
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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.
M38-10
(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.
M38-10
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19 Aug 2022, 08:15
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GMAT CLUB Official Explanation:
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 one 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.
Notice that, from this statement we can get 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 one 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 from (1) that the units digit of \(k\) is 0. For \(\sqrt{k}\) 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 \(10^2=100\)). But from (2) we know that the tens digit of \(k\) is NOT 0, thus \(\sqrt{k}\) is NOT and integer. Sufficient.
Answer: C
General Discussion
14 Jul 2022, 08:24
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The questions asks whether k is a perfect square or not.
St 1: k can be translated as the LCM and multiples of LCM of (2,3,5 & 7) i.e. 210, 420, 630 and so on.
Here, k will only be an integer for all even powers of 2,3,5 & 7. Since we don't know exact value. Insufficient
St 2: Since the factors of a prime number are 1 and the number itself k can be 16, 25, 26... as we don't know the exact
value of k. Insufficient
St 1 & 2 together : Statement 1 tells us that k is perfect square only when the powers of 2,5,3,7 are even, hence the minimum powers of 2&5 is '2' and the number will have 2 trailing zeroes( or 4,6,8..). Now, St 2 tells us that the tens digit has to be either 1 or any single digit prime number, therefore it cannot be '0', hence even powers of 2&5 are not possible. All the remaining numbers with odd powers will not be perfect squares. Sufficient
Answer : C
St 1: k can be translated as the LCM and multiples of LCM of (2,3,5 & 7) i.e. 210, 420, 630 and so on.
Here, k will only be an integer for all even powers of 2,3,5 & 7. Since we don't know exact value. Insufficient
St 2: Since the factors of a prime number are 1 and the number itself k can be 16, 25, 26... as we don't know the exact
value of k. Insufficient
St 1 & 2 together : Statement 1 tells us that k is perfect square only when the powers of 2,5,3,7 are even, hence the minimum powers of 2&5 is '2' and the number will have 2 trailing zeroes( or 4,6,8..). Now, St 2 tells us that the tens digit has to be either 1 or any single digit prime number, therefore it cannot be '0', hence even powers of 2&5 are not possible. All the remaining numbers with odd powers will not be perfect squares. Sufficient
Answer : C
