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Updated on: 25 Sep 2025, 10:03
Originally posted by Sajjad1994 on 08 Sep 2025, 01:24.
Last edited by Sajjad1994 on 25 Sep 2025, 10:03, edited 2 times in total.
Last edited by Sajjad1994 on 25 Sep 2025, 10:03, edited 2 times in total.
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For which of the following integers, \(n\), is the number \(60n\) equal to the square of another integer?
Indicate all such values of n.
\(A. 2^{28}\)
\(B. 2^9 × 3^5\)
\(C. 3^3 × 5^5\)
\(D. 2^{10} × 3^9 × 5^8\)
\(E. 3 × 5 × 7^{28}\)
Indicate all such values of n.
\(A. 2^{28}\)
\(B. 2^9 × 3^5\)
\(C. 3^3 × 5^5\)
\(D. 2^{10} × 3^9 × 5^8\)
\(E. 3 × 5 × 7^{28}\)
Updated on: 25 Sep 2025, 11:01
Originally posted by Sajjad1994 on 25 Sep 2025, 10:32.
Last edited by Sajjad1994 on 25 Sep 2025, 11:01, edited 1 time in total.
Last edited by Sajjad1994 on 25 Sep 2025, 11:01, edited 1 time in total.
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Official Explanation
First factor the number 60 into its prime factors. Since 60 = 4 × 15, we have \(60 = 2^2 × 3 × 5.\) Notice that 2 appears squared in this expression, but that 3 and 5 appear to the first power. For a number to be the square of another integer, each prime in the prime factorization of the number must be raised to an even power.
(A) is not correct, because \(2^{28} × 60\) will equal \(2^{30} × 3 × 5,\) which cannot be the square of an integer, because 3 and 5 are still to the first power.
(B) is not correct, because multiplying 60 by \(2^{9} × 3^{5}\) leaves an expression with \(5^1\).
(C) is correct, because \((2^2 × 3 × 5) × (3^3 × 5^5) = 2^2 × 3^4 × 5^6,\) which is the square of an integer, because all prime factors appear to even powers. Explicitly \(2^2 × 3^4 × 5^6\) is equal to the square of \((2^1 × 3^2 × 5^3).\)
(D) is not correct because of the \(5^8\) part, which multiplied by the 5 in \(2^2 × 3 × 5\) yields \(5^9\), which will not be the square of an integer, because the exponent is odd.
In (E), the product of \(3 × 5 × 7^{28}\) and \(2^2 × 3 × 5\) is \(2^{30} × 3^2 × 5^2 × 7^{28},\) which equals \((2^15 × 3^1 × 5^1 × 7^{14})2,\) so (E) is correct. The answer is (C) and (E).
Answer: C,E
First factor the number 60 into its prime factors. Since 60 = 4 × 15, we have \(60 = 2^2 × 3 × 5.\) Notice that 2 appears squared in this expression, but that 3 and 5 appear to the first power. For a number to be the square of another integer, each prime in the prime factorization of the number must be raised to an even power.
(A) is not correct, because \(2^{28} × 60\) will equal \(2^{30} × 3 × 5,\) which cannot be the square of an integer, because 3 and 5 are still to the first power.
(B) is not correct, because multiplying 60 by \(2^{9} × 3^{5}\) leaves an expression with \(5^1\).
(C) is correct, because \((2^2 × 3 × 5) × (3^3 × 5^5) = 2^2 × 3^4 × 5^6,\) which is the square of an integer, because all prime factors appear to even powers. Explicitly \(2^2 × 3^4 × 5^6\) is equal to the square of \((2^1 × 3^2 × 5^3).\)
(D) is not correct because of the \(5^8\) part, which multiplied by the 5 in \(2^2 × 3 × 5\) yields \(5^9\), which will not be the square of an integer, because the exponent is odd.
In (E), the product of \(3 × 5 × 7^{28}\) and \(2^2 × 3 × 5\) is \(2^{30} × 3^2 × 5^2 × 7^{28},\) which equals \((2^15 × 3^1 × 5^1 × 7^{14})2,\) so (E) is correct. The answer is (C) and (E).
Answer: C,E
