# Simplify (use variables) and plot this summation?

GROUPS:
 I'd like to simplify this equation by using variables. For some reason I can't get the right order and way to define them: sum Part[IntegerDigits [b, 10], i+1] * (10^i+10^(Length[IntegerDigits [b, 10]]-1-i)),  i = 0 to Length[IntegerDigits [b, 10]]-1 I have tried this: a=IntegerDigits [b, 10], k=Length[a]-1, sum Part[a, i+1] * (10^i+10^(k-i)),  i = 0 to k but without success.Then for some reason equation doesn't give right solution with free b variable. WolframAlpha claims it is: 11 (b + 10)But with set variable b=153 calculation is correct: 504:https://www.wolframalpha.com/input/?i=b%3D153,+sum+Part%5BIntegerDigits+%5Bb,+10%5D,+i%2B1%5D+*+(10%5Ei%2B10%5E(Length%5BIntegerDigits+%5Bb,+10%5D%5D-1-i)),%C2%A0+i+%3D+0+to+Length%5BIntegerDigits+%5Bb,+10%5D%5D-1Thanks for any help provided, -Marko
 Bill Simpson 1 Vote When b has not been assigned any value then the value of IntegerDigits [b, 10] will be just IntegerDigits [b, 10], the Length of that will be 2, the value of Part[IntegerDigits [b, 10], 1] will be b and the value of Part[IntegerDigits [b, 10], 2] will be 10.That explains why you are having no luck simplifying your expression without having assigned a value to b.Perhaps there is some way of using quotient(b,10^i) and mod(b,10^i) without having assigned any value to b and still be able to uncover some algebraic relationship that will allow some simplification.Either that or try to describe as simply and clearly as absolutely possible what math property it is that you are trying to capture and simplify and someone might be able to offer another way of expressing this to get a simpler form.If I have understood what your sum is trying to accomplish then I think the result might be b + FromDigits[ Reverse[ IntegerDigits[b]]] That is written using Mathematica notation. I can get WolframAlpha to understand individual parts of that, but not if I try to combine the functions. You should check that very carefully to make certain that I have made no mistakes.