Add 1, change nothing

Algebra Level 5

Each of given 100 numbers was increased by 1. Then each number was increased by 1 once more. Given that the first time the sum of the squares of the numbers was not changed as compared to the original sum of squares, find how this sum was changed the second time, i.e., find the absolute difference between the current sum of squares and the original sum of squares.


The answer is 200.

Mehul Arora by Mehul Arora , 20, India

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.

3 solutions

Rishik Jain
Feb 19, 2016

x 1 2 + x 2 2 + x 3 2 + ⋯ + x 100 2 = ( x 1 + 1 ) 2 + ( x 2 + 1 ) 2 + ( x 3 + 1 ) 2 + … ( x 100 + 1 ) 2 x 1 2 + x 2 2 + x 3 2 + ⋯ + x 100 2 = x 1 2 + x 2 2 + x 3 2 + ⋯ + x 100 2 + 2 ( x 1 + x 2 + x 3 + ⋯ + x 100 ) + 100 2 ( x 1 + x 2 + x 3 + ⋯ + x 100 ) + 100 = 0 x 1 + x 2 + x 3 + ⋯ + x 100 = − 50 ( ( x 1 + 2 ) 2 + ( x 2 + 2 ) 2 + ( x 3 + 2 ) 2 + … ( x 100 + 2 ) 2 ) − ( x 1 2 + x 2 2 + x 3 2 + ⋯ + x 100 2 ) = 4 ( x 1 + x 2 + x 3 + ⋯ + x 100 ) + 400 = 4 × ( − 50 ) + 400 200 x_1^2+x_2^2+x_3^2 + \dots + x_{100}^2=(x_1+1)^2+(x_2+1)^2+(x_3+1)^2 + \dots (x_{100}+1)^2 \\ x_1^2+x_2^2+x_3^2 + \dots + x_{100}^2=x_1^2+x_2^2+x_3^2 + \dots + x_{100}^2 + 2(x_1+x_2+x_3+\dots +x_{100})+100 \\ 2(x_1+x_2+x_3+\dots +x_{100})+100=0 \\ x_1+x_2+x_3+\dots +x_{100}=-50 \\ ((x_1+2)^2+(x_2+2)^2+(x_3+2)^2 + \dots (x_{100}+2)^2)-(x_1^2+x_2^2+x_3^2 + \dots + x_{100}^2)=4(x_1+x_2+x_3+\dots +x_{100})+400 \\ =4 \times (-50) +400 \\ \large \boxed{200}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Surya Sharma
Feb 20, 2016

Bonus: The numbers are (-50,-49.....,48,49)

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Yash Mehan
Feb 19, 2016

You add 1. sum of squares does not change. half of one = 0.5 exploit that square of minus numbers is positive. think that can we put all the 100 numbers as -0.5, as square of -0.5 would be equal to the square of 0.5 square of -0.5 = 0.25. sum of all 0.25 = 0.25 *100 = 25. earlier it was also 25, as {(-0.5)^2 } * 100 = 25 now add 1 all of the 0.5. all numbers become 1.5

their square is 2.25.

2.25*100 = 225.

225-25 =200.

there you go

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...