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If x is the smallest positive integer such that 485,100 divided by x i

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Sajjad1994
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If x is the smallest positive integer such that 485,100 divided by x i [#permalink] New post   19 Aug 2019, 10:39
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A
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C
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E

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75% (hard)

Question Stats:

46% (02:23) correct 54% (02:34) wrong based on 61 sessions
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If x is the smallest positive integer such that 485,100 divided by x is twice the square of an integer, then x must be

A. 2

B. 11

C. 22

D. 77

E. 105
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C
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Kinshook
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If x is the smallest positive integer such that 485,100 divided by x i [#permalink] New post   Updated on: 21 Aug 2019, 01:27
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SajjadAhmad wrote:
If x is the smallest positive integer such that 485,100 divided by x is twice the square of an integer, then x must be

A. 2

B. 11

C. 22

D. 77

E. 105


Given: x is the smallest positive integer such that 485,100 divided by x is twice the square of an integer,

Asked: x must be ?

485,100 = 2^2 * 3^2* 5^2* 7^2 * 11 = 2 (2*3^2*5^2*7^2*11)


x =2*11 = 22

IMO C

Originally posted by Kinshook on 19 Aug 2019, 10:46.
Last edited by Kinshook on 21 Aug 2019, 01:27, edited 1 time in total.
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Re: If x is the smallest positive integer such that 485,100 divided by x i [#permalink] New post   19 Aug 2019, 10:54
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SajjadAhmad wrote:
If x is the smallest positive integer such that 485,100 divided by x is twice the square of an integer, then x must be

A. 2

B. 11

C. 22

D. 77

E. 105


\(\frac{485100}{x}\) = 2*\(a^2\)

Prime factorisation of 485100 = \((2^2)*(3^2)*(5^2)*(7^2)*(11)\)

According to the given statement,

\(\frac{(2^2)*(3^2)*(5^2)*(7^2)*(11)}{x}\) = 2*\(a^2\)


For a variable/number to be square, it should have prime factors with even digits of exponents.
Now, taking out a 2 from both the sides:

\(\frac{(2)*(3^2)*(5^2)*(7^2)*(11)}{x}\) = \(a^2\)

and the factors of \(a^2\) will be \((3^2)*(5^2)*(7^2)\)

The remaining numbers are (2)*(11) which can be the smallest positive integer x can be for the equation to be an integer.

Therefore, the answer is C
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Re: If x is the smallest positive integer such that 485,100 divided by x i [#permalink] New post   20 Aug 2019, 11:14
SajjadAhmad wrote:
If x is the smallest positive integer such that 485,100 divided by x is twice the square of an integer, then x must be

A. 2

B. 11

C. 22

D. 77

E. 105


485100 / x = 2y^2

=> 242550/x = y^2
=> y^2 = 25*9*49*22/x

For y^2 to be a perfect square, x has to be 22 (as 25, 9, 49 are perfect squares)

So, answer should be C.
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Re: If x is the smallest positive integer such that 485,100 divided by x i [#permalink] New post   20 Aug 2019, 23:55
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SajjadAhmad wrote:
If x is the smallest positive integer such that 485,100 divided by x is twice the square of an integer, then x must be

A. 2

B. 11

C. 22

D. 77

E. 105


Similar question to practice: https://gmatclub.com/forum/if-y-is-the- ... 10513.html
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Re: If x is the smallest positive integer such that 485,100 divided by x i [#permalink] New post   21 Aug 2019, 01:19
\(485,100=2^2*3^2*5^2*7^2*11\)

\(\frac{485100}{x}=2*(a^2*b^2*c^2)\)

So \(x\) must cancel out any prime with an odd power and the resulting integer must be 2*(primes raised to even powers)

Lets re-write 485100 as \(2*11*2*(3^2*5^2*7^2)\)

If we can cancel out the \(11*2\), we get our required structure [2*(primes raised to even powers)]

Therefore \(x\) must be 11*2=22

Answer is (C)
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Re: If x is the smallest positive integer such that 485,100 divided by x i [#permalink] New post   25 Aug 2019, 18:34
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SajjadAhmad wrote:
If x is the smallest positive integer such that 485,100 divided by x is twice the square of an integer, then x must be

A. 2

B. 11

C. 22

D. 77

E. 105


We can create the equation:

485,100/x = 2k^2

242,550 = x * k^2

Now let’s factor 242,550:

242,550 = 2 * 3^2 * 5^2 * 7^2 * 11 = 22 * (3 * 5 * 7)^2

Since k must be a perfect square, then the largest value that k can be is 3 * 5 * 7. Thus, the smallest value of x is 22.

Answer: C
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Re: If x is the smallest positive integer such that 485,100 divided by x i [#permalink]
25 Aug 2019, 18:34
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