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17 Apr 2021, 17:47
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E
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Difficulty:
95%
(hard)
Question Stats:
48% (02:38) correct
52%
(02:50)
wrong
based on 69
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Let k be the smallest positive integer such that \(\frac{k}{2}\) is a perfect square and \(\frac{k}{3}\) is a perfect cube. Then the prime factorization of k has the form \(2^c\)*\(3^d\) for positive integers c and d. What is the smallest positive integers divisible by both c and d?
A. 1
B. 3
C. 4
D. 6
E. 12
A. 1
B. 3
C. 4
D. 6
E. 12
17 Apr 2021, 20:40
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twobagels
Now \(k=2^c*3^d\)
1) \(\frac{k}{2}=\frac{2^c3^d}{2}=2^{c-1}3^d=x^2\), as it is given to be a perfect square.
What does this tell us about c and d?
c: c-1 is even, so c is odd.
d: d is even.
2) \(\frac{k}{3}=\frac{2^c3^d}{3}=2^{c}3^{d-1}=y^3\), as it is given to be a perfect cube.
What does this tell us about c and d?
c: c is multiple of 3.
d: d-1 is multiple of 3.
When we combine the two we get
c: c is odd and multiple of 3, so c can be 3, 9, 15 and so on. Minimum value =3
d: d-1 is odd and multiple of 3, so d-1 can be 3, 9, 15 and so on. Thus d can be 4, 10, 16 and so on. Minimum value =4
Now the question in its wordings is basically looking for the LCM(3,4), which is 12.
E
ramlala
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17 Apr 2021, 20:44
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twobagels
Minimum value of C=3 and d=4 to satisfy above conditions as \(\frac{k}{2}\) is a perfect square and \(\frac{k}{3}\) is a perfect cube.
So samllest positive integer divisible by both c and d (LSM) = 12
IMO E
