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Re: The integer k is equal to m^2 for some integer [#permalink]
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Fail2Success wrote:
KarunMendiratta wrote:
Carcass wrote:
The integer ๐‘˜ is equal to \(๐‘š^2\) for some integer ๐‘š. If ๐‘˜ is divisible by 20 and 24, what is the smallest possible positive value of ๐‘˜?

Show: :: OA
3600


\(k = m^2\)
\(20 = (2^2)(5)\)
\(24 = (2^3)(3)\)

I. \(m^2\) is divisible by 20
i.e. \(m^2\) must have at-least two 2s and one 5

II. \(m^2\) is divisible by 24
i.e. \(m^2\) must have at-least three 2s and one 3

This means, \(m^2\) must have at-least three 2s, one 3 and one 5 in it!

Since, \(k\) is a perfect square here, each prime factor must have even power
i.e. \(k = (2^4)(3^2)(5^2) = 3600\)



am sorry to ask this , how can we confirm it as perfect square ? and one more question is why dont we can have (2^2)(3^2)(5^2)


Since, The integer ๐‘˜ is equal to \(๐‘š^2\) for some integer ๐‘š.
A perfect square is a number that can be expressed as the product of two equal integers. This is why ๐‘˜ must be a perfect square.

If we take ๐‘˜ as \((2^2)(3^2)(5^2) = 900\), it is not divisible by 24
\(\frac{900}{24} = 37.5\) not an integer
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Re: The integer k is equal to m^2 for some integer [#permalink]
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