For a positive integer \(N\), let \(S(N)\) be the sum of digits in the decimal representation of \(N\). Let \(F(N)\) be the sum of all the numbers obtained by erasing one or more numbers from the right end of \(N\).

Prove that \[S(N)+9F(N)=N\]

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## Comments

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TopNewestAt first let \(N=\displaystyle\sum_{k=0}^n a_k 10^k\). We have \(S(N)=\displaystyle \sum_{k=0}^n a_k\). Now we calculate and simplify \(F(N)\). A little observation is required to find the compact form of \(F(N)\)

\[F(N) = \sum_{i=1}^n \sum_{j=i}^n a_j 10^{j-i}=\sum_{i=1}^{n} \sum_{j=0}^{i-1} a_{i} 10^j=\sum_{i=1}^n\left(a_{i}\sum_{j=0}^{i-1} 10^j\right)=\sum_{i=1}^n a_{i}\dfrac{10^i-1}{9}.\]

The rest is straight. Using the last form of \(F(N)\) we have

\[9F(N)=\sum_{i=1}^n a_i\left(10^i -1\right)=\sum_{i=1}^n \left(a_i10^i-a_i\right)=\sum_{i=1}^n a_i10^i -\sum_{i=1}^n a_i =N-S(N)\]

and the result follows. \(\square\)

NOTE: If you're wondering how I reformed the sum of \(F(N)\) in the third step above, say \(N=3265\). Then clearly\[F(3625)=326+32+3.\]

This is the form of the first sum. Now break further into

\[\begin{eqnarray*} F(3625) &=&326+32+3 \\ &=&(300+20+6)+(30+2)+3 \\ &=&(300+30+3)+(20+2)+6 \\ &=&3(10^2+10^1+10^0)+2(10^1+10^0)+6\cdot 10^0. \end{eqnarray*}\]

Observe the number of occurrences of the powers of \(10\) for the \(n\)th digit and generalize to get the second sum.

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Great job!

Another way of doing it is inducting on the number of digits of \(N\).

Clearly the statement is true for any one digit-number.

Assume that the statement is true for any \(k\)-digit number.

Now for any \((k+1)\)- digit number \(b\), let \(b=10a+c\) where \(a\) is a \(k\)-digit number and \(c\) is a one-digit number.

Notice that, \(S(b)=S(a)+c\) and \(F(b)=F(a)+a\)

So, \(S(b)+9F(b)=S(a)+c+9F(a)+9a\).

From the induction hypothesis, \(S(a)+9F(a)=a\). Substituting that, we get,

\(S(b)+9F(b)=10a+c=b\).

So the result holds for all positive integers \(N\).

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BTW, someone's downvoting every damn comment he's passing by. -_-

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At first it seems rather tedious. However, if we prove by induction on the number of digits the calculations will magically disappear!

When number of digits is 1, \(F(N)=0, S(N)=n\), done.

Suppose it was true for all \(k\) digit numbers. Consider a \((k+1)\) digit number \(X\) with last digit \(m\). Let \(Y\) be the \(k\) digit number without \(m\). Then \(X=10Y+m\). Hence \(S(X)+9F(X)=S(Y)+m+9F(Y)+9Y\) (since we missed out \(Y\) when calcualting \(F(Y)\)).

But by inductive hypothesis, \(S(Y)+9F(Y)=Y\), hence \(S(X)+9F(X)=Y+m+9Y=10Y+m=X\) and we are done.

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Sorry, I did not see Mursalin's comment below. I just realized it was a repetition.

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Interesting.

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