Units problem

Number Theory Level 2

Find the units digits of the number $1^{2}$ + $2^{2}$ + $3^{2}$ +........................+ $99^{2}$

For more problems ,click here .

The answer is 0.

5 solutions

Rama Devi
May 26, 2015

Using n (n+1)(2n+1)/6,we get the answer as 0.

Akash Deep
Jun 21, 2014

(1^2 + 2^2 + 3^2 + ......... + 10^2 ) unit digit we can find out very easily that is 45. now further what we observe is that 11 - 20 squares addition will also have the same unit digit and so on for any 10 no;s from ............1 to ...............0 will have same unit digit sum = 45 now uptill 100 there will be 10 such like (1 - 10 )to (91 -100) and thus unit digit of 45 * 10 = 0 and thus unit digit is 0

Vishwesh Agrawal
Apr 13, 2015

Hint use n (n+1)(2n+1)/6

Sudoku Subbu
Jan 17, 2015

when we consider 1 - 10 numbers we get unit digit as 1+4+9+6+5+6+9+4+1+0= 45 ; when we consider 11-20 numbers also we get as 45 because we consider only unit digit in 11-20 unit digits are 1-10. therefore for all numbers unit digit is 45 X 10 = 450 therefore the unit digit is 0 . [you may ask that iam considering till hundred because unit digit of 100^{2} is 0 ]

Sunil Pradhan
Jul 6, 2014

sum of squares from 1 to n = n(n+1)(2n+1)/6 = 99 × 100 × 199/6

out of this 99 × 100 = 990 × 10 (990 divisible by 6)

unit digit = 0

Units problem

For more problems ,click here .

The answer is 0.

5 solutions

0 pending reports