Wondering how you would solve this problem.

Jomo

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The fraction a/b is equals to the repeating decimal 0.ababab. If a and b are single-digit positive integers, what is the value of a + b.
What I did was say that (10a+b)/99 = a/b and then b(10a+b) = 99a.
Since the one's place on the lhs comes from b2 I realized that a can only be 1, 4, 5, 6 and/or 9.
Then by trial and error I found a=b=9. Of course after seeing the answer I realized that I should have seen that one immediately but probably not any others (if in fact there were any).

In any case I would like to see how others would have solved this.
Thanks,
Steven
 
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MarkFL

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I agree with your approach, up to here:

\(\displaystyle \frac{10a+b}{99}=\frac{a}{b}\)

This leads to the quadratic in \(b\):

\(\displaystyle b^2+10ab-99a=0\)

Thus (discarding the negative root):

\(\displaystyle b=\frac{-10a+\sqrt{100a^2+396a}}{2}=-5a+\sqrt{a(25a+99)}\)

Now, the discriminant must be a perfect square, and so we require:

\(\displaystyle a=n_1^2\)

\(\displaystyle 25a+99=n_2^2\)

where \(n_i\in\mathbb{N}\)

Subtracting the former from the latter, we obtain:

\(\displaystyle 2^3\cdot3a+3^2\cdot11=(n_2+n_1)(n_2-n_1)\)

This means that in order for the LHS to be factorable over the integers, and for \(a\) to be a single digit natural number, \(a\) must either be 3 or 9, but \(a\) must be a perfect square, so \(a\) must be 9, and so:

\(\displaystyle b=-5\cdot9+\sqrt{9(25\cdot9+99)}=-45+9\sqrt{25+11}=54-45=9\)

Thus:

\(\displaystyle (a,b)=(9,9)\)
 

Jomo

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I agree with your approach, up to here:

\(\displaystyle \frac{10a+b}{99}=\frac{a}{b}\)

This leads to the quadratic in \(b\):

\(\displaystyle b^2+10ab-99a=0\)

Thus (discarding the negative root):

\(\displaystyle b=\frac{-10a+\sqrt{100a^2+396a}}{2}=-5a+\sqrt{a(25a+99)}\)

Now, the discriminant must be a perfect square, and so we require:

\(\displaystyle a=n_1^2\)

\(\displaystyle 25a+99=n_2^2\)

where \(n_i\in\mathbb{N}\)

Subtracting the former from the latter, we obtain:

\(\displaystyle 2^3\cdot3a+3^2\cdot11=(n_2+n_1)(n_2-n_1)\)

This means that in order for the LHS to be factorable over the integers, and for \(a\) to be a single digit natural number, \(a\) must either be 3 or 9, but \(a\) must be a perfect square, so \(a\) must be 9, and so:

\(\displaystyle b=-5\cdot9+\sqrt{9(25\cdot9+99)}=-45+9\sqrt{25+11}=54-45=9\)

Thus:

\(\displaystyle (a,b)=(9,9)\)
I actually tried that technique but gave up on it after solving for b using the quadratic formula.

I am lost with what you did however. What exactly did you mean by This means that in order for the LHS to be factorable over the integers, and for a to be a single digit natural number, a must either be 3 or 9, but a must be a perfect square, so a must be 9...? Why just 3 or 9??

Edit: Or did you mean to say a can be 3, 6 or 9?
 

MarkFL

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Yes, \(a\) could also be 6 (I missed that...it can be any positive single digit multiple of 3), but it still needs to be a perfect square, and so 9 is still the only option. :)
 

Jomo

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OK, thanks Mark.
For some reason it is very hard for me to figure out what someone meant to say but I am improving with that. Now I understand what you said perfectly.
Eloquently done!
 

Otis

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To be honest, I might have failed to realize that symbols a and b could represent the same digit (but I'm not sure because of case a=9).

Anyway, I would have focused entirely on b and simply used division rules.

a/1, a/2, a/4, a/5, a/8 terminate
a/3, a/6 repeat 3 or 6
a/7 eventually repeats 142857
a/9 repeats a

Where a/b reduces, the same rules apply to the reduced ratio.

Again, I'm not sure whether I would have answered 'no solution' or realized the possibility of 0.aaaaaa…=0.ababab… when dividing by 9 (thus answering 18).

Yer a trickster, Jomo!

;)
 

MarkFL

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OK, thanks Mark.
For some reason it is very hard for me to figure out what someone meant to say but I am improving with that. Now I understand what you said perfectly.
Eloquently done!
If it's something I've posted, and it's not clear or seems to be wrong, please don't hesitate to ask. I make typos, and I make mistakes, despite my efforts to minimize that. :)
 

Jomo

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If it's something I've posted, and it's not clear or seems to be wrong, please don't hesitate to ask. I make typos, and I make mistakes, despite my efforts to minimize that. :)
I did just that. That is what this forum is for! Thanks!!
 

Jomo

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Otis

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This is the way the problem was given to me.
And you passed it on, as is!

No, seriously, there's nothing wrong with the wording. The issue is with me; sometimes, I'm not all there.

🕶
 

Jomo

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And you passed it on, as is!

No, seriously, there's nothing wrong with the wording. The issue is with me; sometimes, I'm not all there.

🕶
That's better than me, as I am never all there.
 
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