# Unusual algebra problem (having trouble understanding the explanation)

#### m3dicat3d

##### New member
Hi all, new to the forum here. I posted this about a week ago on a different web board and had difficulty understanding the reply. I was hoping I might be able to ask someone here if they can help me make sense of this... I'm amazed at how much trouble I'm having with it...

The Problem:

If 18*sqrt(18) = r*sqrt(t), where t and r are positive integers and r > t, which of the following could be the value of r*t? (solution is 108 but I have no clue how this war arrived at)

The Response:

since everything in sight is positive we can square both sides without fear.

thus from:

18√18 = r√t, we have:

(324)(18) = r2t

183 = 5832 = r2t.

clearly t < 18, or else r2t > t3 > 5832. so t is some divisor of 18: 1,2,3,6,or 9.

if t = 1, r = √(5832), which is not an integer.
if t = 2, r = √(2916) = 54 <--this works ( (18)3/2 = (9)(18)2, which has square root 3*18 = 54).
if t = 3, r = √(1944), not an integer
if t = 6, r = √(972), not an integer
if t = 9, r = √(648), not an integer

(look at the prime factorization of 18 cubed)

so the only case where r and t are integers with r > t is t = 2, r = 54, hence rt = 108.

My follow up to the response:

"since everything in sight is positive we can square both sides without fear.

thus from:

√18 = r√t, we have:

(324)(18) = r2t

183 = 5832 = r2t.

clearly t < 18, or else r2t > t3 > 5832. so t is some divisor of 18: 1,2,3,6,or 9."

I think where I'm getting hung up in the explanation is the "clearly t < 18, or else r2t > t3 > 5832. so t is some divisor of 18: 1,2,3,6,or 9."

I understand that if t > r that the result would be > 5832. The part I'm having trouble with is how to indentify that t is in fact less than 18 specifically.How do we know this? And how does the responder know to use a factor/divisor of 18 and not some other integer?

Is there a link/website or some section of math one would practice in a book/class for a problem of this nature? It's important to me to understand the "why" of it instead of memorizing "how" to do a particular problem type.

Thanks.

#### Sudharaka

##### Well-known member
MHB Math Helper
I think where I'm getting hung up in the explanation is the "clearly t < 18, or else r2t > t3 > 5832. so t is some divisor of 18: 1,2,3,6,or 9."

I understand that if t > r that the result would be > 5832. The part I'm having trouble with is how to indentify that t is in fact less than 18 specifically.How do we know this?
Hi m3dicat3d, Welcome to MHB! Since $$r>t$$ we have, $$r^2>t^2$$. Multiplying both sides by $$t$$ we get, $$r^2 t>t^3$$. Now suppose that $$t>18$$. Then,

$r^2 t>t^3>18^3=5832$

which is a contradiction (since we have shown $$r^2 t=5832$$). Therefore $$t<18$$.

Kind Regards,
Sudharaka.

Last edited:

#### awkward

##### Member
Hi m3dicat3d, Welcome to MHB! Since $$r>t$$ we have, $$r^2>t^2$$. Multiplying both sides by $$t$$ we get, $$r^2 t>t^3$$. Now suppose that $$t>18$$. Then,

$r^2 t>t^3>18^3=5832$

which is a contradiction (since we have shown $$r^2 t=5832$$). Therefore $$t<18$$.

Since, $$r^2 t=18^3$$ we have, $$t\mid 18^3$$. Suppose $$t\nmid 18$$. Then, the only possibility is $$t>18 (t=18^2\mbox{ or }18^3)$$ which is a contradiction (we have shown that $$t<18$$).

Kind Regards,
Sudharaka.
Hi Sudharka,

Frankly, I find that last step unclear.
Since, $$r^2 t=18^3$$ we have, $$t\mid 18^3$$. Suppose $$t\nmid 18$$. Then, the only possibility is $$t>18 (t=18^2\mbox{ or }18^3)$$ which is a contradiction (we have shown that $$t<18$$).
What if $$t = 4$$, for example? Then $$t \mid 18^3$$, but $$t \nmid 18$$, yet $$t < 18$$.

Here is a suggestion that I think will work. We have $$r^2 t = 2^3 3^6$$, so the prime factors of $$r$$ and $$t$$ must be among 2 and 3. Let $$r = 2^a 3^b$$, where it is possible that $$a = 0$$ or $$b = 0$$. Then $$r^2 = 2^{2a} 3^{2b}$$, so $$t = 2^{3-2a} 3^{6-2b}$$. What are the possibilities?

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi Sudharka,

Frankly, I find that last step unclear.

What if $$t = 4$$, for example? Then $$t \mid 18^3$$, but $$t \nmid 18$$, yet $$t < 18$$.
I am sorry but my reasoning for that part is incorrect (hence edited my previous post). I will let you know if I come up with anything useful.

#### soroban

##### Well-known member
Hello, m3dicat3d!

If $18\sqrt{18} = r\sqrt{t}$, where $t$ and $r$ are positive integers and $r > t$,
which of the following could be the value of $r\!\cdot\!t$?

Note that: .$18\sqrt{18} \:=\:18\sqrt{9\!\cdot\!2} \:=\:18\!\cdot\!\sqrt{9}\!\cdot\!\sqrt{2} \:=\:18\!\cdot\!3\!\cdot\!\sqrt{2} \:=\:54\sqrt{2}$

So we have: .$54\sqrt{2} \:=\:r\sqrt{t}$

Therefore: .$\begin{Bmatrix}r &=& 54 \\ t &=& 2 \end{Bmatrix} \quad\Rightarrow\quad r\!\cdot\!t \:=\:54\!\cdot\!2 \:=\:108$

#### Opalg

##### MHB Oldtimer
Staff member
There is also another solution, because $54\sqrt2 = (27\times2)\sqrt2 = 27\sqrt4\sqrt2 = 27\sqrt8$. So you could take $r=27$, $t=8$. That also satisfies $r>t$, but this time $rt = 27\times8 = 216.$ Therefore $216$ is a solution.

#### Bacterius

##### Well-known member
MHB Math Helper
If 18*sqrt(18) = r*sqrt(t), where t and r are positive integers and r > t, which of the following could be the value of r*t? (solution is 108 but I have no clue how this war arrived at)
Here is a somewhat brute force solution:

$$18 \sqrt{18} = r \sqrt{t}$$

$$18^3 = r^2 t$$

$$2^3 \times 3^6 = r^2 t$$

Find all possible values of $r^2$ such that $r$ is an integer:

$$r^2 = \{ 2^2 | 3^2 | 3^4 | 2^2 3^2 | 2^2 3^4 | 3^6 | 2^2 3^6 \} ~ \implies ~ r = \{ 2^1 | 3^1 | 3^2 | 2^1 3^1 | 2^1 3^2 | 3^3 | 2^1 3^3 \}$$

Find the corresponding values of $t$:

$$t = \{ 2^1 3^6 | 2^3 3^4 | 2^3 3^2 | 2^1 3^4 | 2^1 3^2 | 2^3 | 2^1 \}$$

Now keep all the $(r, t)$ pairs which satisfy $r > t$, and we have only two solutions left:

$$(r, t) = (2^1 3^3, 2^1), (3^3, 2^3)$$

In conclusion, $r t = 2^1 3^3 2^1 = 108$ and $r t = 3^3 2^3 = 216$.

PS: the bars I used are not standard notation but I thought they were more readable than commas in this case. I wanted to make a LaTeX table but it didn't work for some reason.

#### Deveno

##### Well-known member
MHB Math Scholar
Hi Sudharka,

Frankly, I find that last step unclear.

What if $$t = 4$$, for example? Then $$t \mid 18^3$$, but $$t \nmid 18$$, yet $$t < 18$$.

Here is a suggestion that I think will work. We have $$r^2 t = 2^3 3^6$$, so the prime factors of $$r$$ and $$t$$ must be among 2 and 3. Let $$r = 2^a 3^b$$, where it is possible that $$a = 0$$ or $$b = 0$$. Then $$r^2 = 2^{2a} 3^{2b}$$, so $$t = 2^{3-2a} 3^{6-2b}$$. What are the possibilities?
^
this. outstanding. this limits it to 8 possibilities:

r = 1, t = 5832 (a = 0, b= 0). but r < t.
r = 3, t = 648 (a = 0, b= 1). but r < t.
r = 9, t = 72 (a = 0, b= 2). again...
r = 27, t = 8 (a = 0, b= 3) <--we have a winner!
r = 2, t = 1458 (a = 1, b= 0). but r < t.
r = 6, t = 972 (a = 1, b= 1).
r = 18, t = 18 (a = 1, b= 2). oh so close....
r = 54, t = 2 (a = 1,b = 3) <---another winner!

for the two winners, we have rt = 108, and rt = 216.

#### m3dicat3d

##### New member
To everyone that has responded, thanks SOOOO much. Especially to Denevo, who was in fact the repsonder from the Math Help Forum board where I asked this question initially.

I'll be honest and say that soroban's explaination makes the most sense to me.

Believe it or not, this is a question from an SAT prep manual that I'm attempting to help someone get ready for. I taught HS math in AZ for 3 years under a temp certificate, and being a degree holding "geologist" and not a mathematician, HS math was fairly easy for me to teach. I mention this only because I have never once seen a problem like this and I'm quite frankly amazed at how puzzling I'm finding it.

I'm curious about two things... First off, what does the vertical symbol in something like "then t *vertical symbol* 183" mean as it's being used here (I've only seen that in set-builder notation and not how it is being used here)?

The other thing, and I alluded to this in the initial post already, is what "category" of algebra does a problem like this fall under? For instance, someone might see a problem and say there's not enough info to solve it, but once they discover the theorem behind it, it makes total sense. I'd like to look into this more on the conceptual level.

I'm actually studying for my own certification exam here in Texas to teach HS math again under a full certification, and as embarassed as I am to admit that this SAT problem is causing me so much confusion, it's still important for me to understand it, so that I can explain the conceptual basis to the student I'm attempting to help out (which is how I've taught math in the past--teach them "why" instead of "how").

Thanks so very much again to all of you, I genuinely appreciate so many people offering their input into this #### topsquark

##### Well-known member
MHB Math Helper
I wouldn't worry about the SAT question. I have a MS in Physics and there are problems in the Introductory texts that can take me. You can do what you can do. Next time you see something like this you now know how to do it. -Dan

#### caffeinemachine

##### Well-known member
MHB Math Scholar
I'm curious about two things... First off, what does the vertical symbol in something like "then t *vertical symbol* 183" mean as it's being used here (I've only seen that in set-builder notation and not how it is being used here)?
$t|18^3$ is read as $t$ divides $18^3$. More generally, in number theory, $a|b$ is read as $a$ divides $b$.

Edit: In post#7 Bacterius doesn't intend to use $|$ as 'divides'.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
The other thing, and I alluded to this in the initial post already, is what "category" of algebra does a problem like this fall under? For instance, someone might see a problem and say there's not enough info to solve it, but once they discover the theorem behind it, it makes total sense. I'd like to look into this more on the conceptual level.
This was a basic 'Number Theory' problem. If you have time, you can read David M. Burton's 'Elementary Number Theory'. For such problems you would just need to read the first three chapters. For a first reading, I think, the first 7 chapters are fairly readable.