The sum of all the digits of the numbers from | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To find the sum of all digits of numbers from $1$ to $100$,we consider numbers from $00$ to $99$ (since $00$ to $09$ are the same as $0$ to $9$ an

Vedclass · Accepted Answer

$901$

Vedclass · Answer

(C) To find the sum of all digits of numbers from $1$ to $100$,we consider numbers from $00$ to $99$ (since $00$ to $09$ are the same as $0$ to $9$ and $100$ contributes $1+0+0=1$).There are $100$ numbers,each having $2$ digits,making a total of $200$ digit positions.Each digit ($0$ through $9$) appears an equal number of times in the units and tens places.Number of times each digit appears $= \frac{200}{10} = 20$ times.Sum of digits from $0$ to $9 = 0+1+2+3+4+5+6+7+8+9 = 45$.Total sum of digits for numbers $00$ to $99 = 20 	imes 45 = 900$.Finally,adding the sum of digits of $100$ $(1+0+0 = 1)$:Total sum $= 900 + 1 = 901$.

The sum of all the digits of the numbers from $1$ to $100$ is

Similar Questions

In a $G.P.$,the first term is $7$,the last term is $448$,and the sum is $889$. Find the common ratio.

Determine the $25^{th}$ term of an $A.P.$ whose $9^{th}$ term is $-6$ and the common difference is $\frac{5}{4}$.

If the sum of $n$ terms of a $G.P.$ is $255$,the $n^{th}$ term is $128$,and the common ratio is $2$,then the first term will be:

Given $x = 1 + a + a^2 + ... \infty$ $(a < 1)$ and $y = 1 + b + b^2 + ... \infty$ $(b < 1)$,find the value of $1 + ab + a^2b^2 + ... \infty$.

If the $10^{th}$ term of a geometric progression is $9$ and the $4^{th}$ term is $4$,then its $7^{th}$ term is

Vedclass Test Series

Exam Paper Generator

Online Exam Module