# Wondering how you would solve this problem.

#### Jomo

##### Elite Member
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

Last edited:

#### MarkFL

##### Super Moderator
Staff member
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:

$$\displaystyle (a,b)=(9,9)$$

• jonah2.0 and topsquark

#### Jomo

##### Elite Member
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

#### MarkFL

##### Super Moderator
Staff member
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

##### Elite Member
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

##### Senior Member
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

#### MarkFL

##### Super Moderator
Staff member
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. • topsquark

#### Jomo

##### Elite Member
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

##### Elite Member
Yer a trickster, Jomo! This is the way the problem was given to me.

#### Otis

##### Senior Member
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

##### Elite Member
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.