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!
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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:
Excellent! Upvoted.
Thanks
Thank you. That was really fun.
Looking forward to the next one :)
I love such riddles. :)
Here is my attempt to solve it:
a = 1, b = 2, c = 4
12 / 21 = 24 / 42
That's good - gave you a small upvote :-)
but there are more solutions...
7.75x7.75/7.75x7.75=7.75x7.75/7.75x7.75
Thanks, but 7.75 is not a 2-digit number.
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