Finding an exponent variable.

**Cooper** · November 24th 2011, 06:06 PM

I've gone through this thing 12 times, and not once found a viable answer without a calculator, as asked. I don't want to know the answer as much as I want to know how to go about, so help would be appreciated.

The product of the factors of 6^16 can be expressed as 6^k, find k.

**takatok** · November 24th 2011, 07:17 PM

Hint: Say 2 numbers are factors of $6^{16}$ . If you multiply them together, by their very definition they = $6^{16}$

So if $6^{16}$ only had 6 factors. We could just divide by 2 to get 3 pairs of $6^{16}$ . Then the answer would be $3*6^{16}$ Obviously this isn't the actual number of factors since there woudn't be a 3 out front. the answer will most likely be in $a*6^{16}$ where a = multiple of 6.

So now you just need to know how many factors 6^16 has. Can dig around for that or reply back here, and I can let you know. There is a pretty quick way to tell using prime factorization.

**Cooper** · November 29th 2011, 04:32 PM

Thank you! A few questions though; wouldn't it be (6^16)^a? Since we're looking for the product? Also, how would I go about finding the the number of factors 6^16 has? I mean, if this were a question on a test, I don't think I'd have the time to list them all out and count them. My best guess would be to list them out as a bunch of 2's and 3's, but I'm not sure if that's correct.

**Soroban** · November 29th 2011, 06:07 PM

Hello, Cooper!

I did some scribbling and I think I've solved it . . . maybe not.

$\text{The product of the factors of }N \:=\:6^{16}\text{ can be expressed as }6^k. \;\text{ Find }k.$

First, we know that:. $N \:=\:2^{16}\cdot 3^{16}$

The factors of $2^{16}$ are:. $A \:=\:\{2^0,\:2^1,\:2^3\:\hdots\: 2^{16}\}$

The factors of $3^{16}$ are:. $B \:=\:\{3^0,\:3^1,\:3^3,\:\hdots\:3^{16}\}$

If we pair every element of $A$ with every element of $B,$
. . we will have all the factors of $N.$

Start the list . . .

$(2^03^0)\cdot(2^03^1)\cdot(2^03^2)\;\cdots\;(2^03^ {16})$
. . Their product is:. $(2^0)^{17}(3^{(0+1+2+3+\cdots+16)}) \;=\;2^{(0\,\cdot17)}\cdot3^{136}$

$(2^13^0)\cdot(2^13^1)\cdot(2^13^2)\:\cdots\;(2^13^ {16})$
. . Their product is:. $(2^1)^{17}(3^{(0+1+2+3+\cdots+16)}) \;=\;2^{(1\,\cdot17)}\cdot3^{136}$

$(2^23^0)\cdot(2^23^1)\cdot(2^23^2)\:\cdots\;(2^23^ {16))$
. . Their product is:. $(2^2)^{17}(3^{(0+1+2+3+\cdots+16)}) \;=\;2^{(2\,\cdot17)}\cdot3^{136}$

. . . . . . $\vdots$ . . . . . . . . . . . . . $\vdots$ . . . . . . . . . . . . . . . $\vdots$

$(2^{16}3^0),\:(2^{16}3^1),\:(2^{16}3^3),\;\hdots\; (2^{16}3^{16})$
. . Their product is:. $(2^{16})^{17}(3^{(0+1+2+3+\cdots+16)}) \;=\;2^{(16\,\cdot17)}\cdot3^{136}$

Multiplying, we have:. $(2^{17})^{(0+1+2+\cdots+16)}\cdot(3^{136})^{17}$

Therefore, the product is: . $N \;=\;2^{(17\cdot136)}\cdot3^{(136\cdot17)} \;=\; 2^{952}\cdot3^{952} \;=\;6^{952}$

**takatok** · November 29th 2011, 10:52 PM

First off you are right it should have been 6^16^a. Not sure what I was thinking

Well to answer your question about the number of factors:
For any number N: if you take its prime factorization (that is break it down into multiplication by only primes), it will look like this:

$p_0^{a_0}\cdot p_1^{a_1}\cdot p_2^{a_2}\cdot ...p_n^{a_n}$

then its number of factors (including 1 and N) will be:

$(a_0+1)\cdot(a_1+1)\cdot(a_2+1)\cdot ...(a_n+1)$

This is because each prime factor $p_i$ can take a value from $p_i^0\ to\ p_i^{a_i}$ and be combined with each other in all possible ways.

So for $6^{16} = 2^{16}\cdot 3^{16}$

(16+1)*(16+1) = 289 factors.

So each pair of these factors will multiply and equal 6^16. (But wait there is an odd number!.. thats because $2^8\cdot 2^8 = 2^{16}$ so we only count it once.

So ${(6^{16})}^{289/2} = 6^{2312}$

Soraban proved this empirically by enumerating every possilbility then multiplying. He just made a slight multiplication error at the end. 136*17=2312, not 952!

(I think he accidentally multiplied by 7)

So he came out with the same answer as above. However, since now you know how to calculate all the factors for a number, you have no need to list them all out for really big numbers if its ever the case.

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