Question

Square all the integers from 1 to 10 inclusive. Then, round each number to the nearest hundred. Finally, sum the numbers. What do you get?

Answer 1

We get the sum of numbers rounded off to nearest 100 = 300

Explanation:

Integers from 1 to 10 inclusive.

Squaring them:

`1^(2) = 1\\2^(2) = 4\\3^(2) = 9\\4^(2) = 16\\5^(2) = 25\\6^(2) = 36\\7^(2) = 49\\8^(2) = 64\\9^(2) = 81\\10^(2) = 100`

Rounding each of them to the nearest 100:

All the number less than 50 are rounded off to previous 100, which is 0.

All the other numbers i.e. 64, 81 are rounded off to 100.

100 is already rounded off, we do not need to round it off.

`1 \rightarrow 0 \\4\rightarrow 0\\9\rightarrow 0\\16\rightarrow 0\\ 25\rightarrow 0\\36\rightarrow 0\\49\rightarrow 0\\64\rightarrow 100\\81\rightarrow 100\\`

Now, taking the sum of the rounded off numbers:

`0+0+0+0+0+0+0+100+100+100 = 300`

We get the sum of numbers rounded off to nearest 100 = 300

Calculating actual sum of squares from 1 to 10:

Using the formula:

`S_n = (n(n+1)(2n+1))/(6)`

Here n = 10

`1^2+2^2+3^2+..... + 10^2 = (10 * 11 * 21)/(6) \\\Rightarrow \bold {385}`

And sum of rounded off numbers = 300