Digit at the unit place of the sum of (1 !)2+(2 !)2+(3 !)2 ldots ldots ldots +(2008 !)2 is

Question

Digit at the unit place of the sum of (1 !)2+(2 !)2+(3 !)2 ldots ldots ldots +(2008 !)2 is (A) 5 (B) 7 (C) 9 (D) 6

Tardigrade Admin · Accepted Answer

Given the sum S = (. 1 ! .)2 + (. 2 ! .)2 + (. 3 ! .)2 + ....... + (. 2008 ! .)2 Since the digit at units place is zero in n ! for n ≥ 5 Hence, S = (. 1 .)2 + (. 2 .)2 + (. 6 .)2 + (. 24 .)2 + numbers having zero at units place = 617 + all other numbers having zero at units place

Q. Digit at the unit place of the sum of $(1!)^{2} + (2!)^{2} + (3!)^{2} \dots\dots\dots + (2008!)^{2}$ is

5

7

9

6

Given the sum $S = (1!)^{2} + (2!)^{2} + (3!)^{2} + ....... + (2008!)^{2}$
Since the digit at units place is zero in $n!$ for $n \geq 5$
Hence, $S = (1)^{2} + (2)^{2} + (6)^{2} + (24)^{2} +$ numbers having zero at units place
$= 617 +$ all other numbers having zero at units place

Q. Digit at the unit place of the sum of (1!)2+(2!)2+(3!)2………+(2008!)2 is

5

7

9

6

Given the sum S=(1!)2+(2!)2+(3!)2+.......+(2008!)2 Since the digit at units place is zero in n! for n≥5 Hence, S=(1)2+(2)2+(6)2+(24)2+ numbers having zero at units place =617+ all other numbers having zero at units place

Q. Digit at the unit place of the sum of $(1!)^{2} + (2!)^{2} + (3!)^{2} \dots\dots\dots + (2008!)^{2}$ is

Given the sum $S = (1!)^{2} + (2!)^{2} + (3!)^{2} + ....... + (2008!)^{2}$
Since the digit at units place is zero in $n!$ for $n \geq 5$
Hence, $S = (1)^{2} + (2)^{2} + (6)^{2} + (24)^{2} +$ numbers having zero at units place
$= 617 +$ all other numbers having zero at units place