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Maths Puzzle
find the unit place digit of (1!)^3+(2!)^3+(3!)^3+...+(100!)^3
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- 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.
- 10 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 - 10 years agoHelpfull: Yes(0) No(0)
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