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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#Algebra #DigitSum #NumberBaseRepresentation #ProofTechniques #Induction

Easy Math Editor

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Comments

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

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

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

Jubayer Nirjhor - 6 years, 9 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 $N$ .

Mursalin Habib - 6 years, 9 months ago

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

Jubayer Nirjhor - 6 years, 9 months ago

@Jubayer Nirjhor – I don't mind being downvoted.

Mursalin Habib - 6 years, 9 months ago

@Mursalin Habib – Neither do I. But I do mind being spammed.

Jubayer Nirjhor - 6 years, 9 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.

Joel Tan - 6 years, 9 months ago

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

Joel Tan - 6 years, 9 months ago

Interesting.

Shubham Agrawal - 6 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$