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14 Jul 2022, 08:00
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C
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42% (02:04) 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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Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
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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
14 Jul 2022, 09:03
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From st. 1 we have, k is a multiple of every single-digit prime number
Thus k = 2 * 3 * 5 * 7 * constant (c) = 210 * c
If √k is an integer then we need the constant term c in k to be
c = (2 * 3 * 5 * 7)^(odd powers)
But we are not given any information about c. So this statement is not sufficient.
From st. 2 we have, the tens digit of k is a factor of a single digit prime number
So tens digit of k could be = 1, 2, 3, 5, 7
If k = 25 or 36 then answer is yes. But if not then No.
So So this statement is not sufficient
Combining St 1 and St 2 we have
If √k is an integer then k must at least be 210 * 210 = 44100. Thus we see that units digit and tens digit of k must have 0 in order to √k to be an integer. But st 2 clearly says that the tens digit is not 0. Hence √k is not an integer.
So answer choice (C)
Thus k = 2 * 3 * 5 * 7 * constant (c) = 210 * c
If √k is an integer then we need the constant term c in k to be
c = (2 * 3 * 5 * 7)^(odd powers)
But we are not given any information about c. So this statement is not sufficient.
From st. 2 we have, the tens digit of k is a factor of a single digit prime number
So tens digit of k could be = 1, 2, 3, 5, 7
If k = 25 or 36 then answer is yes. But if not then No.
So So this statement is not sufficient
Combining St 1 and St 2 we have
If √k is an integer then k must at least be 210 * 210 = 44100. Thus we see that units digit and tens digit of k must have 0 in order to √k to be an integer. But st 2 clearly says that the tens digit is not 0. Hence √k is not an integer.
So answer choice (C)
