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Brainsteem Mathematics Challenges: Fractions by rycharde

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· @rycharde · (edited)
$19.30
Brainsteem Mathematics Challenges: Fractions
<center>https://i.imgur.com/j3Z2J6H.png</center>

### Question

Let pq be a 2-digit decimal number, for example, if p=3 and q=1 then the number is 31.

Find the possible values of a, b and c, such that

ab/ba = bc/cb

and the fractions are not equal to 1.


---

As always, these questions are designed to be done by hand, without computational assistance. In this case, you should find some structure to the puzzle so that you can then generate many other such triangles of primes.

---

The first correct answer and further interesting comments will be rewarded with an upvote.

Enjoy!

 
---

<center>- - ![](https://i.imgur.com/MtKq9KV.jpg) - - ![](https://i.imgur.com/9Mvn9fs.jpg) - -
Please Comment, Resteem and Upvote. Thanks!

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vote details (41)
@bloodkratos ·
$0.09
I love such riddles. :)

Here is my attempt to solve it:

a = 1, b = 2, c = 4

12 / 21 = 24 / 42
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@rycharde ·
$0.25
That's good - gave you a small upvote :-)
but there are more solutions...
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@buildawhale ·
re-rycharde-brainsteem-mathematics-challenges-fractions-20180227t122217863z
You got a 3.67% upvote from @buildawhale courtesy of @rycharde!
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@fadhlan86 ·
7.75x7.75/7.75x7.75=7.75x7.75/7.75x7.75
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@rycharde ·
Thanks, but 7.75 is not a 2-digit number.
👍  
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vote details (1)
@sadekj ·
$0.54
Thank you, @rycharde. This is interesting. I will try:

1- Based on the assumption ab = (10a+b). 
Similarly bc=(10b+c), and so on.

2- Therefore (10a+b)/(10b+a)= (10b+c)/(10c+b)
=> (10a+b) * (10c+b) = (10b+a) * (10b+c)

3- Now we multiply the brackets to get:
(100 a * c ) + (10 a * b) + (10 b * c) + b^2 = 100 b^2 + (10 b * c) + (10 a * b) + (a * c)
=> 100ac + b^2 = 100b^2+ ac
=> 99 ac = 99 b^2
=> a*c = b^2

4- We know that a,b, and c are different single digit numbers. So, we can easily try what number would fit this easy formula: **ac = b^2**

5- Possible solutions: 
- a = 1, b = 2,  c =4 (because 2^2 = 1*4)
- a = 4, b = 2,  c =1 (because 2^2 = 4*1)
- a = 2, b = 4,  c= 8 (because 4^2 = 2*8)
- a = 8, b = 4,  c= 2 (because 4^2 = 8*2)
- a = 9, b = 6,  c= 4 (because 6^2 = 9*4)
- a = 4, b = 6,  c= 9 (because 6^2 = 4*9)
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vote details (4)
@rycharde ·
$0.54
Excellent! Upvoted.
Thanks
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@sadekj ·
$0.03
Thank you. That was really fun.
Looking forward to the next one :)
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