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Re M3105
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07 Jun 2015, 08:53
Official Solution: T is the set of all positive integers \(x\) such that \(x^2\) is a multiple of both 27 and 375. Which of the following integers must be a divisor of every integer \(x\) in T? I. 9 II. 15 III. 27
A. I only B. II only C. I and II only D. I and III only E. I, II, and III The least common multiple of \(27 = 3^3\) and \(375 = 3*5^3\) is \(3^3*5^3\). For an integer \(x\), the least value of \(x^2\) which is a multiple of \(3^3*5^3\) is \(3^4*5^4\), so the least value of \(x\) is \(3^2*5^2\). Therefore, \(T = \{3^2*5^2;\) \(2*(3^2*5^2);\) \(3*(3^2*5^2);\) \(4*(3^2*5^2); ...\}\). The question asks which of the options must be a divisor of every integer \(x\) in T. So, the answer is I and II only (\(27 = 3^3\) is NOT a divisor of \(3^2*5^2\)). Answer: C
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Re: M3105
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17 Nov 2015, 08:54
Hi Bunuel Can you please tell what is the difference between this an that problem? I am confused/ Looks like I am missing somethin. Thanks. If we are told that x is integer, and x^2 is divisible by 27, should not x^2 for sure have 4 factors of 3? ifnisapositiveintegerandn2isdivisibleby72then129929.html
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17 Nov 2015, 10:08



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Re: M3105
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16 Jan 2017, 03:03
Bunuel wrote: Official Solution:
T is the set of all positive integers \(x\) such that \(x^2\) is a multiple of both 27 and 375. Which of the following integers must be a divisor of every integer \(x\) in T? I. 9 II. 15 III. 27
A. I only B. II only C. I and II only D. I and III only E. I, II, and III
The least common multiple of \(27 = 3^3\) and \(375 = 3*5^3\) is \(3^3*5^3\). For an integer \(x\), the least value of \(x^2\) which is a multiple of \(3^3*5^3\) is \(3^4*5^4\), so the least value of \(x\) is \(3^2*5^2\). Therefore, \(T = \{3^2*5^2;\) \(2*(3^2*5^2);\) \(3*(3^2*5^2);\) \(4*(3^2*5^2); ...\}\). The question asks which of the options must be a divisor of every integer \(x\) in T. So, the answer is I and II only (\(27 = 3^3\) is NOT a divisor of \(3^2*5^2\)).
Answer: C beautifully explained just to add we choosed 3^4*5^4[/m] as 3^3*5^3 will have even number of factors (3+1)(3+1)..



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Re: M3105
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10 Feb 2017, 02:16
Shouldn't it be for the integer X, the least value of X^2 which is a multiple of 3^3∗5^3 is 3^4∗5^3 ?



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10 Feb 2017, 02:22



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22 Aug 2017, 14:14
I think this is a highquality question and I agree with explanation.



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03 Apr 2018, 14:30
Hi Bunuel
I have a doubt here. How come multiple of x^2 is x^4? Shouldn't it be 2*x^2 ? Please explain this concept to me. Thank you



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03 Apr 2018, 20:58



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06 May 2018, 03:23
Hi Bunuel,
Why does the value of x^2 have to be 3^4*5^4? if we assume x^2 is 3^6*5^6 then even 27 will be a divisor or x



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07 May 2018, 00:37



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24 Sep 2018, 13:39
T=(15, 30, 45,60...............)
Only 15 is the ans. Why are people including x^2 in the set.



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24 Sep 2018, 20:44










