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# Proof Problem Of The Day - Digits and Sums!

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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Note by Mursalin Habib
3 years, 3 months ago

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At 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.

- 3 years, 3 months ago

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 $$N$$.

- 3 years, 3 months ago

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

- 3 years, 3 months ago

I don't mind being downvoted.

- 3 years, 3 months ago

Neither do I. But I do mind being spammed.

- 3 years, 3 months ago

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.

- 3 years, 3 months ago

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

- 3 years, 3 months ago

Interesting.

- 3 years, 2 months ago