If R = 1! + 2! + 3! …. 199!, what is the units digit of R?

Solution

• \(1! = 1 \)hence unit’s digit of \(1! = 1\)
• So, units’ digit of 2! = unit’s digit of\( 1! *2 = 1*2 =2\)
• Units’ digit of 3! = units’ digit of \((2! *3)\) = unit’s digit of \((2*3) = 6\)
• Unit’s digit of 4! = units’ digit of \((3! *4) =\) unit’s digit of \((6*4) = 4 \)
• Unit’s digits of 5! = unit’s digit of \((4! *5) =\) unit’s digit of \((4*5) = 0\)
• So, Unit’s digit of 6! = unit’s digit of \((5! *6) =\) unit’s digit of \((0*6) = 0 \)
• Thus, unit’s digit of all the numbers from 5! To 199! will be 0.
• Hence, unit’s digit of R = (unit’s digit of 1! ) + (unit's digit of 2!) + (Unit's digit of 3!) + (Unit’s digit of 4!)

o Or, unit’s digit of R = unit’s digit of \((1 +2 +6 +4) = 3\)

Alternate solution,

• We know that 5! = 1*2*3*4*5

o Since, it contains 2*5 unit’s digit must be 0.
o Hence for all n! where \(n ≥ 5\) and n is positive integer, unit’s digit will be 0.
• Thus, unit’s digit of R = Unit’s digit of \((1! +2! +3! +4!)\) = Unit’s digit of \((1+2+6+24 ) = 3\)