Find the least positive integer m such that 2^2.3^5.7.11.m is a perfect square.

[Fatimamalik]

From this 2010 thread:

Suppose that in standard factored form a = p1^e1 * p2^e2 ...* pk^ek, where k is a positive integer; p1, p2,...,pk are prime numbers; and e1, e2,...,ek, are positive integers.

a) What is the standerd factored form for a^2?

I know the answer to this is p1^2*e1 * p2^2*e2... * pk^2*ek.

b) Find the least posivtive integer n such that 2^5 * 3 * 5^2 * 7^3 * n is a perfect square. Write the resulting product as a perfect square.

n = 42, 2^5 * 3 * 5^2 * 7^3 * n = 5880^2

The answer to this is in the back of the book....

For the same question there is a part c for which im confused

C. Find the least positive integer m such that 2².3⁵.7.11.m is a perfect square. Write the resulting product as a perfect square

 

[ksdhart]

First off, I'm assuming the decimal points are intended to be multiplication symbols, and thus it's actually: 2³ * 3⁵ * 7 * 11 * m. If that assumption is correct, then think about what the problem is asking. You have several constants and a variable. What if you multiplied all of the constants together? And then you know you want to find an m such that the expression equals a perfect square. Well, a perfect square can be written as some integer n squared, right? So, what if you take the square root of both sides?

$\displaystyle 149688m=n^2$

$\displaystyle \sqrt{149688m}=n$

You know that both m and n have to be integers, so what is the smallest m that fulfills those conditions?

 

[pka]

From this 2010 thread:


For the same question there is a part c for which im confused

C. Find the least positive integer m such that 2².3⁵.7.11.m is a perfect square. Write the resulting product as a perfect square

$\displaystyle \large m=3\cdot 7\cdot 11$ WHY?

 

[Ishuda]

From this 2010 thread:


For the same question there is a part c for which im confused

C. Find the least positive integer m such that 2².3⁵.7.11.m is a perfect square. Write the resulting product as a perfect square

One way to look at questions such as this [perfect square, perfect cube, etc.] is to consider just what it means and the prime factorization of the number. For example, for squares, a is a perfect square if a^1/2 is an integer (implying also that a is an integer). That means there is some integer b such that
b² = a
Now if b is factored into prime factors p₁, p₂, p₃, ... then
b = $\displaystyle p_1^{e_1}\, *\, p_2^{e_2}\, *\, p_3^{e_3}\, ...,\, *\, p_n^{e_n}$
where the e_j are integers, or squaring b,
b² = $\displaystyle p_1^{2\, e_1}\, *\, p_2^{2\, e_2}\, *\, p_3^{2\, e_3}\, ...,\, *\, p_n^{2\, e_n}$.
Thus all the exponents of the prime decomposition of a must be even.

Also note that integer in this context means a positive integer.