silly factors

[nanase]

please help me out on this :

Someone found the product of the first six multiples of three.
3 X 6 X 9 X 12 X 15 X 18 = 524880

How many factors of 524880 are the square of an integer greater than 1?

being an average student in maths i tried this below

I listed all the square numbers until 20 (20²=400)
I also realise that there is connection between factor of 12 which is 3, and factor of 18 which is 9, which if I multiply will give 27, so 27² will be an answer also. (is my assumption correct?)
thus 36 squared will also be a factor of 524880, because 36 is made up of 9 and 4, factor from 12 and 18. Is my reasoning valid?
pls add and correct and improve my reasoning? how do i get all the factors from this question? anything I missed out?
many thanks

 

[lev888]

What does a square look like? N*N. So, what
we are counting is how many ways are there to use the available numbers to make up N, so that we still have the exact same numbers for the second N.
I would try this approach:
List all prime factors of the first six multiples of three.
Divide then into 2 equal sets. There is only one 5, so it can be safely discarded.
Take one set and count how many different products your can make.

 

[nanase]

What does a square look like? N*N. So, what
we are counting is how many ways are there to use the available numbers to make up N, so that we still have the exact same numbers for the second N.
I would try this approach:
List all prime factors of the first six multiples of three.
Divide then into 2 equal sets. There is only one 5, so it can be safely discarded.
Take one set and count how many different products your can make.

Hai sir I am still wondering with your last sentences.
what do you mean with divide into 2 equal sets?
I obtained 2, 3, 5 as prime factors from the numbers

 

[nanase]

the answer key is 14.
I'm still depressed in figuring how to find those 14 numbers? I could only find 9 in my workings

i tried listing the prime factors of the first six multiples of 3
3 = 3
6 = 2 x 3
9 = 3²
12 = 2² x 3
15 = 3 x 5
18 = 2 x 3²

I'm curious how these can help me find the squared factors?

 

[nanase]

2
2.2
2.3
2.2.3
2.2.3.3
2.2.3.3.3
2.2.3.3.3.3
3
3.3
3.3.3
3.3.3.3
3.2 (repetitive)
3.2.2 (repetitive)
3.3.2
3.3.2.2 (repetitive)
3.3.3.2
3.3.3.2.2 (repetitive)
3.3.3.3.2
3.3.3.3.2.2 (repetitive)

anyway still did what you asked me, and excluding the ones with repetitive, I obtained 14 possible answers. I am not sure if this is the valid approach though. I also want to ask from the mathematical perspective:
1. why the 5 is omitted?
2. why multiplying the factors that exclude 5, when I square the results, will still be the factor of 524880? what is the mathematical reasoning behind this?
3. from the prime factors you asked me to list, I counted 8 of threes, 4 of twos and 1 of 5. you asked me to divide into 2 equal sets, why ?

thanks , I am just... intrigued

 

[Otis]

… I obtained 2, 3, 5 as prime factors from the numbers

Hello. lev888 is talking about grouping the prime factors of 524880:

2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 5

Those are the factors available to make squares, and as lev888 noted we can ignore 5 because we can't make 5² with only one factor of 5.

Here are some of the squares we can make:

(2)(2)
(2×2)(2×2)
(3)(3)
(3×3)(3×3)
(2×3)(2×3)
(2×2×3)(2×2×3)
(2×2×3×3)(2×2×3×3)

There are 14 arrangements, before you run out of factors.

?

 

[Otis]

Ah, we just cross-posted. Glad you figured it out.

\(\;\)

 

[Otis]

… why multiplying the factors that exclude 5, when I square the results, will still be the factor of 524880? …

I'm not sure I understand that question, but I'll guess. The number we sqaure comes from prime factors of 524880. Squaring doesn't change the factors; it just doubles their presence.

For example, let's look at the factor 2×3×3
If we square it, we have 2×2×3×3×3×3

524880 = 2×2×2×2×3×3×3×3×3×3×3×3×5

In this example, I started with 2×3×3, but I could have started with any of the 14 possibilities. In each case, after squaring the factor, I would end up with a "part" of 524880's prime factorization. In other words, each squared number must be a factor of 524880 because it's a part of 524880.

\(\;\)

 

[nanase]

I'm not sure I understand that question, but I'll guess. The number we sqaure comes from prime factors of 524880. Squaring doesn't change the factors; it just doubles their presence.

For example, let's look at the factor 2×3×3
If we square it, we have 2×2×3×3×3×3

524880 = 2×2×2×2×3×3×3×3×3×3×3×3×5

In this example, I started with 2×3×3, but I could have started with any of the 14 possibilities. In each case, after squaring the factor, I would end up with a "part" of 524880's prime factorization. In other words, each squared number must be a factor of 524880 because it's a part of 524880.

\(\;\)

gosh thanks! you understood and explained my questions well, I guess that also answers why lev888 asked me to divide into 2 sets, so i can just focus with the possible arrangements from 2 2 3 3 3 3 and ignore the 5 ! thanks a lot bro! I understand how it works now and feel happier ! haha

 

[pka]

Someone found the product of the first six multiples of three.
3 X 6 X 9 X 12 X 15 X 18 = 524880 How many factors of 524880 are the square of an integer greater than 1?

\(\displaystyle 524880=2^4\cdot3^8\cdot 5\) as you said we can forget the five.
Any square factor, \(\displaystyle >1\), looks like \(\displaystyle 2^a\cdot 3^b\) where \(\displaystyle (a,b)\in \{0,2,4\}\times\{0,2,4,6,8\},~(a,b)\ne(0.0)\).
There are fourteen such possible pairs.

 

[Jagulba]

3 X 6 X 9 X 12 X 15 X 18

Lets count power of 2,3,5 in each terms in above multiplication

2	0+1+0+2+0+1=4
3	1+1+2+1+1+2=8
5	0+0+0+0+1+0=1

here you see we only have 1 power of 5 which means we can’t make a square number which has 5 in it, we need at least 2.
now 2 we can because we have 4
we also can consider 4 since its separate integer and we have 2⁴ = 4²

for 3 we have 3,9, 81 and don’t forget 27


beside these numbers 2,4,3,9,27,81 as I call them originals we can also have combination of 2s and 3s
2*3,2*9,2*27,2*81

4*3,4*9,4*27,4*81
simply

(2 numbers of power of 2 )*(4 numbers of powers of 3) + original numbers = 14

 

[pka]

This is an interesting topic is counting.
How many divisors does four million five hundred and thirty six thousand have.
To answer that first factor \(\displaystyle 4536000=2^6\cdot 3^4\cdot 5^3\cdot 7\) SEE HERE
Each divisor looks like \(\displaystyle 2^a\cdot 3^b\cdot 5^c\cdot 7^d\) where \(\displaystyle 0\le a\le 6,~0\le b\le 4,~0\le c\le 3,~\&~0\le d\le 1\)
So that means there are \(\displaystyle 7\cdot 5\cdot 4\cdot 2=280\) divisors.
How, many of those are perfect squares? Well the exponents must be even:
\(\displaystyle a\in\{0,2,4,6\},~b\in\{0,2,4\},~c\in\{0,2\},~d\in\{0\}\) or \(\displaystyle 4\cdot 3\cdot 2\cdot 1=24\)

Here is a fun question. If \(\displaystyle N=7^9\cdot 5^7\cdot 3^8\cdot 2^6\) then \(\displaystyle N\) ends in how many zeros?
If \(\displaystyle N=7^a\cdot 5^b\cdot 3^c\cdot 2^d\) then \(\displaystyle N\) ends in how many zeros? [a,b,c,~\&~d are non-negative integers].

 

[Jagulba]

oh yeah this was a cool discovery the first time i saw it. you don't want to repeat "the answer" 3 times though
now if log2=0.301 , log6=0.778 and log21=1.322, then what is the number of digits of N?

 

[nanase]

Here is a fun question. If \(\displaystyle N=7^9\cdot 5^7\cdot 3^8\cdot 2^6\) then \(\displaystyle N\) ends in how many zeros?
If \(\displaystyle N=7^a\cdot 5^b\cdot 3^c\cdot 2^d\) then \(\displaystyle N\) ends in how many zeros? [a,b,c,~\&~d are non-negative integers].

give me some hint! i am interested in solving this

 

[pka]

Here is a fun question. If \(\displaystyle N=7^9\cdot 5^7\cdot 3^8\cdot 2^6\) then \(\displaystyle N\) ends in how many zeros? If \(\displaystyle M=7^a\cdot 5^b\cdot 3^c\cdot 2^d\) then \(\displaystyle M\) ends in how many zeros? [a,b,c,~\&~d are non-negative integers]./QUOTE]

Given N above it ends in six zeros. Think how a number ends in zeros. The number \(\displaystyle 2^{100}\cdot 7^{50} \) ends in no zeros.​

The number \(\displaystyle 3^{100}\cdot 5^{50} \) ends in no zeros The number \(\displaystyle 2^{45}\cdot 5^{44} \) ends in 44 zeros​

The number \(\displaystyle 2^{45}\cdot 5^{46} \) ends in 45 zeros.​

If \(\displaystyle M=7^a\cdot 5^b\cdot 3^c\cdot 2^d\) then \(\displaystyle M\) ends in how many zeros?​

 

[nanase]

im working on your hint,
so if 2⁴⁵.5⁴⁶ = 10⁴⁵.5
so this is how you get the number of 45 zeros.

but when the power is in a,b,c,d, then the number of zeros will be b-d or d-b?

 

[pka]

im working on your hint,
so if 2⁴⁵.5⁴⁶ = 10⁴⁵.5
so this is how you get the number of 45 zeros.
but when the power is in a,b,c,d, then the number of zeros will be b-d or d-b?

Actually the expected answer was If \(\displaystyle \min\{b,d\}\)
However, in one class that was a really talented kid who gave a Correct answer: \(\displaystyle \frac{|b+d|-|b-d|}{2}\), of course that is \(\displaystyle \min\{b,d\}\).