self Maths Puzzle

find the unit place digit of (1!)^3+(2!)^3+(3!)^3+...+(100!)^3

Read Solution (Total 2)

3 is the last I.e. unit digit. as 1!,2!,3! & 4! will participate in giving unit digit after that only 0 will be the unit digit, so just add 1+2+6+4=13,so 3 is the unit digit.
11 years agoHelpfull: Yes(0) No(2)

ans=9
1!=1 ; 2!=2 ; 3!=6 ; 4!=24; after that all no.s having '0' at its unit place.
now consider only unit place,(neglecting zeros, as it doesn't affect the addition)
1^3 + 2^3 + 6^3 + 4^3 = 1+8+216+64 = 289 => unit place = 9
11 years agoHelpfull: Yes(0) No(0)

self Other Question

no of zeroes at the end of the expression (2!)^2!+4!^4!+8!^8!+9!^9!+10!^10!+11!^11! Roy is now 4 year older than Erik and 2 years older than Iris.If in 2 years ,Roy will be twice as old as Erik,then in 2 year what would be Roy's age multiplied by Iris's age?