Interesting pattern...

Rani wrote down the numbers 1 - 100 on a piece of paper and then correctly added up all the individual digits of the numbers.

What sum did she obtain?

576 5050 1001 901 2501

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

4 solutions

Consider a 10x10 grid containing values 0 to 99, with 00 in the upper left corner, and 99 in the bottom right. Each row contains values with the same tens digit, Each column contains values with the same ones digit. Notice, 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 0+1+2+3+4+5+6+7+8+9 = 45 .
Notice there is symmetry in that the sum of the ones digits across any row is always 45, and the sum of the tens digits across any column is also 45. Taking the sums separately gives 45 × 10 + 45 × 10 = 2 × 45 × 10 = 900 45\times10 + 45\times10 =2\times45\times10= 900 since 100 was left out of this grid you may add one to account for its absence. giving 901 901

Naren Bhandari
Aug 11, 2018

Let d n m d_{n|m} denote the sum of digits of number from n n to m m .We can notice that the sum of digits of the number follows in following pattern.

d 0 10 = 9 0 + 45 + 1 d 11 20 = 9 1 + 45 + 2 d 21 30 = 9 2 + 45 + 3 = d 90 100 = 9 9 + 45 + 1 c c c since 1 = 1 + 0 + 0 d 0 100 = 9 j = 1 9 j + 45 10 + j = 0 9 j + 1 = 9 45 + 45 10 + 45 + 1 = 45 ( 9 + 10 + 1 ) + 1 = 901 \begin{aligned} d_{0|10} & = 9\cdot 0 + 45 + 1 \\ d_{11|20} & = 9\cdot 1 + 45 +2 \\ d_{21|30} & = 9\cdot 2 + 45 + 3\\ \vdots \ & = \quad \vdots \qquad \vdots \\ d_{90|100} & = 9\cdot 9 +45 + 1\phantom{ccc}\text{since}{\color{#3D99F6} 1=1+0+0 } \\ \hline d_{0|100} & =9\sum_{j=1}^{9} j + 45\cdot 10 + \sum_{j=0}^{9}j +1\\ & = 9\cdot 45 +45\cdot 10 +45+1\\ & = 45\,(9+10+1) +1 = \boxed{901}\end{aligned}

Manifold M
Oct 21, 2018

Let S n S_n denote the sum of the first n n digits. Observe that: 11 + 12 + . . . 19 = ( 10 + 1 ) + ( 10 + 2 ) + . . . + ( 10 + 9 ) 11 + 12 + ... 19 = (10 +1) + (10 + 2) + ... + (10 + 9)

Hence, we compute S 9 S_9 10 times, and adding each multiple of 10, 10 times. So we get the following equation: 10 S 9 + ( 10 10 + 10 20... + 10 90 ) = 450 + 100 ( S 9 ) = 450 + 450 = 900 10 * S_9 + (10*10 + 10*20... + 10*90) = 450 + 100(S_9)= 450 + 450 = 900 Considering the 1 in 100, we get 900 + 1 = 901 900 + 1 = 901 .

Brian Moehring
Aug 11, 2018

Every digit d { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } d \in \{0,1,2,3,4,5,6,7,8,9\} appears in the numbers 00 - 99 00\text{-}99 equally often (note that I write the leading zero if necessary, so it is 06 06 , not 6 6 , but this doesn't change the digit sum). Since there are 100 100 numbers, each with 2 2 digits, and there are 10 10 choices for d d , we see that every d d appears 100 2 10 = 20 \frac{100\cdot 2}{10} = 20 times.

Therefore the sum of the digits 00 - 99 00\text{-}99 , or equivalently 1 - 99 1\text{-}99 , is 20 ( 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) = 20 45 = 900 20 \cdot (0+1+2+3+4+5+6+7+8+9) = 20 \cdot 45 = 900

Then just add the digit sum from the number 100 100 to get an answer of 900 + 1 + 0 + 0 = 901 900 + 1 + 0 + 0 = \boxed{901}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...