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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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To solve this, let's find the smallest k such that k = m^2 and k is divisible by both 20 and 24. Start by finding the least common multiple (LCM) of 20 and 24, which will be the smallest k that satisfies the divisibility requirements.

1. Factorize 20 and 24:
20 = 2^2 * 5
24 = 2^3 * 3

2. Find the LCM:
- The LCM will include each prime factor raised to its highest power across both factorizations:
- LCM(20, 24) = 2^3 * 3 * 5 = 120

3. Since k = m^2 and m must be an integer, k must be a perfect square. The smallest perfect square that is at least 120 and divisible by both 20 and 24 is 3600:

For k = m^2 and k to be a perfect square, every prime factor in its factorization must appear to an even power. From the step above, the minimal factorization needed is 2^3 * 3 * 5 . To make each of these powers even, adjust them to the next even power (since the square of any integer will have even powers for all primes):
Increase 2^3 to 2^4 (next even power)
Increase 3 to 3^2
Increase 5 to 5^2

4. Multiply these adjusted factors:

2^4 * 3^2 * 5^2 = 16 * 9 * 25 = 3600.

If you did not know the rule of exponents and perfect squares, or another shortcut, at step 3, you may have spent a long time crunching tough numbers. Or even aborted the mission at that point. Itโ€™s a hard question.

You need to know these rules. The Ultimate GRE Cheat Sheet has excellent chapters on exponent rules and divisibility. I strongly recommend you get a copy.

With 160 pages of rules and patterns, the cheat sheet is much more comprehensive than the lists of formulas and rules that you will see elsewhere.
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Re: The integer k is equal to m^2 for some integer [#permalink]
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