Solving a Math Problem: Number Interchange & 9 More

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The discussion revolves around a math problem involving a two-digit number where interchanging the digits results in a number that is 9 greater than the original. The equations derived from the problem indicate that the second digit is one more than the first digit, leading to the conclusion that multiple solutions exist. The participant concludes that the number 12 is one valid solution, but acknowledges that any number where the second digit is one more than the first (like 23, 34, etc.) also satisfies the conditions. This indicates that the problem is underdetermined, as there are infinitely many valid combinations. Ultimately, the key takeaway is that there isn't a unique solution, but rather a set of numbers that fit the criteria.
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Let us consider a two digit number. If the digits are interchanged, the new number is 9 more than the original number. The second digit of the original number is one more than the first digit. Find the number.
Let x be the first digit and y be the second digit

(10y + x) - (10x +y) = 9
and
y -x =1

(10y + x) - (10x +y) = 9
9y -9x = 9
y - x = 1

I find out that both equations are the same does that mean that there isn't enough information or have I done something wrong?
From messing around I find the number is 12
 
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It is underdetermined. Any number where y-x = 1 will work.
 
U can't find two separate value OR unique Values

Generally the digits will be x at units place and 1+x at ten's digit and 0<x<9
 
What do you mean "From messing around I find the number is 12"? Just how messy were you? Did you think about 23? 34? 45? 56? 67? 89? All those number have exactly the properties you require. (Of course, that is exactly what Moose352 said: "Any number where y-x = 1 will work."
 
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