Question 232959
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The fact that the sum of the digits is odd and that when you divide the number by 2 the remainder is 1 are both fascinating facts but are not particularly useful in solving the problem.


The fact integer division by 10 has a remainder of 3 guarantees that the last or units digit is 3.  Let *[tex \Large t] represent the tens digit, and let *[tex \Large h] represent the hundreds digit.


The sum of the units digit and the tens digit, *[tex \Large t\ +\ 3] is (meaning equals) one less than the hundreds digit. *[tex \Large h\ -\ 1], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ +\ 3\ =\ h\ -\ 1]


Add 1 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ t\ +\ 4]


The units digit is half the sum of the hundreds (you said hundredths digit, but there cannot be a hundreths digit because dividing the 3 digit number by 10 resulted in an integer, not a decimal fraction.) digit and the tens digit.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 3\ =\ \frac{t\ +\ h}{2}\ \rightarrow\ t\ +\ h\ = 6]


Substitute the expression developed for *[tex \Large h]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ +\ (t\ + 4)\ = 6]


Solve for *[tex \Large t] and then use that value to solve for *[tex \Large h]  Then you can reassemble the number by placing the digits in the proper order.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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