MHB How Can Reversing Digits Triple the Tens Place?

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The problem involves finding a two-digit number where the sum of its digits equals 6, and reversing the digits results in a number that is three times the original tens digit. The equations derived are t + u = 6 and 10u + t = 3t, leading to the conclusion that t = 5u. By substituting these values, it is determined that u = 1 and t = 5, resulting in the original number being 51. Thus, the solution confirms that the original number is 51.
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the sum of the digits of a two digit number is 6, If the digits are reversed, the new number is tree times the original tens number. find the original number.

well just playing with the numbers I got 51 as the original number since 15 is 3 times 5
but doing the problem with equations ?

I tried

$t + u = 6$
$3u = 3t$
but this not got it.

thanks ahead
 
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Hello, karush!

The sum of the digits of a two-digit number is 6.
If the digits are reversed, the new number is three times the original ten's-digit.
Find the original number.
The original number is: $10t + u.$

We are told: .$t + u \,=\,6$ [1]

Also that: .$10u + t \:=\:3t \quad\Rightarrow\quad 10u \,=\,2t \quad\Rightarrow\quad t \,=\,5u$ [2]

Substitute [2] into [1]: .$5u + u \:=\:6 \quad\Rightarrow\quad 6u \,=\,6 \quad\Rightarrow\quad \boxed{u \,=\,1}$

Substiute into [2]: .$t \,=\,5(1) \quad\Rightarrow\quad \boxed{t \,=\,5}$Therefore, the original number is: .$10t + u \:=\:10(5) + 1 \:=\:51$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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