Observation gives rise to a problem!

Number Theory Level 3

$\large \sum_{r=1}^{10^n} r$

What is the value of the sum of digits of the summation above if $n$ is a positive integer?

The answer is 10.

6 solutions

Nihar Mahajan
Apr 12, 2015

When $n=1$ , sum becomes $55 = 2^0(5^1+50^1)$

When $n=2$ , sum becomes $5050 = 2^1(5^2+50^2)$

When $n=3$ , sum becomes $500500 = 2^2(5^3+50^3)$

$\dots \dots$

When $n=n$ , sum becomes $S = 2^{n-1}(5^n+50^n)$ where,

$S$ has $(2n-2)$ zeroes and $two$ $5's$ .

Hence sum of digits $= 5+5=\large\boxed{10}$

Moderator note:

This solution is not complete. You have only shown that it appears to be true for $n=1,2,3$ nor was it practical to manually calculate the sum of the first 100 natural numbers or higher. You should mention the sum of the first $n$ positive integers is equals to $\frac n 2 (n+1)$ then show what Tanishq Varshney did.

one can also see that $\displaystyle \sum_{r=1}^{10^{n}} r=\frac{10^{n}(10^{n}+1)}{2}$

$\large{\frac{10^{2n}}{2}+\frac{10^{n}}{2}}$

Thus we get two $5's$ and the sum is 10

Tanishq Varshney - 6 years, 2 months ago

Can u pls explain why(or maybe how) u said "we get two 5's"..😃 I'm not so good at maths....ill be thankful 😃

Asîf Múktàdîr - 5 years, 7 months ago

10/2=5;100/2=50;1000/2=500

for $10^{100}/2=5\times 10^{99}$ we observe sum of digits is $5$ as rest of the digits are zero.

Hope it helps.

Tanishq Varshney - 5 years, 7 months ago

$CORRECT!!$

Vaibhav Prasad - 6 years, 2 months ago

Love these "CORRECT!" 's lol

David Holcer - 6 years, 2 months ago

hi, can you please explain me how you get that formula 2^(n-1) * (5^n+50^n) I am not good in Mathematics :D thanx

Syed Baqir - 6 years, 2 months ago

$\begin{array}{c}~\displaystyle \sum_{r=1}^{10^n} r &= \dfrac{10^n(10^n+1)}{2}\\~\\ &= \dfrac{10^{2n}+10^n}{2}\\~\\ &= \dfrac{100^n+10^n}{2}\\~\\ &= \dfrac{100^n+10^n}{2^n 2^{1-n}}\\~\\ &= 2^{-(1-n)}\dfrac{100^n+10^n}{2^n}\\~\\ &= 2^{n-1}\dfrac{100^n+10^n}{2^n}\\~\\ &= 2^{n-1}\left[\dfrac{100^n}{2^n}+\dfrac{10^n}{2^n}\right]\\~\\ &= 2^{n-1}\left[\left(\dfrac{100}{2}\right)^n+\left(\dfrac{10}{2}\right)^n\right]\\~\\ &= \boxed{2^{n-1}\left(50^n+5^n\right)}\\~\\ \end{array}$

Micah Wood - 6 years, 2 months ago

thanx for help, means alot

Syed Baqir - 6 years, 2 months ago

Can I ask something?

So you say sum of digits $=10$ for every $n$ .

But what if $n=0$ , which gives the value of $S$ is $1$ ?

Trung Đặng Đoàn Đức - 6 years, 2 months ago

Sorry , I was referring to the 'n' in the given summation of question , where the summation starts from '1'.

Nihar Mahajan - 6 years, 2 months ago

plz check the link given its showing error404 it must be @https://brilliant.org/profile/nihar-pd4fyq/sets/the-problems-that-i-created-with-efforts-and-hard/

Ashwin Upadhyay - 6 years, 2 months ago

@Ashwin Upadhyay – Thanks for spotting!

Nihar Mahajan - 6 years, 2 months ago

@Nihar Mahajan – i didn't understand why you posted the challenge master note on your own solution??? :\

P.S BTW Nice problem & I must say nice observation too :)

& a nice set also

Ashwin Upadhyay - 6 years, 2 months ago

@Ashwin Upadhyay – Lol , its posted by Calvin Sir. Actually , many people are not convenient with sigma notation, hence to make it look easy , I posted it like that (in a pattern).Which is not proved but only intuition. Thanks!

Nihar Mahajan - 6 years, 2 months ago

It was mentioned in the question that n is a positive integer, Trung!!

Archishman Mukherjee - 5 years, 7 months ago

How do you expect us to give answer while you haven't provided the value of 'n'?

Homiee Nanavati - 5 years, 11 months ago

Observe that $\dfrac{1}{1}= 1 \ , \ \dfrac{2}{2}= 1 \ , \ \dfrac{3}{3}= 1 \dots$

So if $n$ is a positive integer and you are asked to find the value of $\dfrac{n}{n}$ , would you ask "How do you expect us to give answer while you haven't provided the value of 'n'?" or would you find the answer as $1$ ?

Nihar Mahajan - 5 years, 11 months ago

A Former Brilliant Member
Apr 19, 2015

Use the identity: 1+2+3+4+...+n = 0.5n(n+1) First evaluate for small n then try to notice a pattern. When n=1, sum is 10(11)/2.= 55 When n=2, sum is 100(101)/2 =550 When n=3, sum is 1000(1001)/2 = 500500 Notice that when we are multiplying n=10^k (k is an integer) by (n+1) the result will be (n+1) followed by k zeros. So it will be of the form 10...00100..00. So 0.5n(n+1) will be 500...000500...00. So there will only be 2 non zero digits in the final sum: 5 and 5. Thus, the value of the sum of digits of the summation above if is a positive integer is 5+5 = 10.

Harish Sasikumar
Nov 17, 2015

$\sum_{r=1}^{10^n}r = \frac{10^n(10^n+1)}{2}=5\times10^{n-1}(10^n+1)$ $=5\times10^{2n-1}+5\times10^{n-1}$ $5000...(2n-1 zeros) + 5000... (n-1)zeros$ $= 5000..5000..$ Hence, the sum of digits = 5+5 = 10 $\forall n\epsilon \{1,2,3,...\}$

Noel Lo
May 21, 2015

You get 50000.........00000500000......00000 so sum is $5+5 = \boxed{10}$

Saumil Joshi
May 21, 2015

I agree with the answer 10 but why not 1 ?

I thought 1 as well. But the question asks for the sum of digits , not the digital root: the digital root is 1, but the sum of digits is certainly 10.

Joel Toms - 5 years, 10 months ago

Aravind Vishnu
Apr 23, 2015

$\text{The sum consists of 2 5's and}\,2(n-1)\,0's. \\ \text{so..............}$

Observation gives rise to a problem!

The answer is 10.

6 solutions

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