 # let n be a natural number and f(n) denotes product of non zero digits in n for example:f(900)=9,f(124)=8 find last digit in ((f(1)+f(2)+f(3).........+f(999))/45 Badiuddin askIITians.ismu Expert
148 Points
12 years ago

Dear puneet

S(1-99)= f(1) +f(2) +.....f(10) + f(11) ......+f(20)+f(21)+ ...................f(99)

= (1+2+3..9) +1 +1*(1+2+..9) + 2 +2*(1+2+3....+9) .......9 + 9*(1+2+3...9)

=(1+2+3....+9)(1+2+3...9) + 2(1+2+3..9)

S(100-199) = (1+2+3..9) +1*1  +1*1*(1+2+..9) + 1*2 +1*2*(1+2+3....+9) .......+1*9 + 1*9*(1+2+3...9)

=1*(1+2+3...9)(1 +2 +3...9)+ 2*1*(1+2+3..9)

S(200-299) = 2*(1+2+3..9) +2*1  +2*1*(1+2+..9) + 2*2 +2*2*(1+2+3....+9) .......+2*9 + 2*9*(1+2+3...9)

=2*(1+2+3...9)(1 +2 +3...9)+ 2*2*(1+2+3..9)

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S(900-999)=9*(1+2+3..9) +9*1  +9*1*(1+2+..9) + 9*2 +9*2*(1+2+3....+9) .......+9*9 +9 *9*(1+2+3...9)

=9*(1+2+3...9)(1 +2 +3...9)+ 9*2*(1+2+3..9)

so

S = (1+2+3...9)2 + 2(1+2+3..9)  + (1+2+3...9)+ 2*(1+2+3..9)2

=(1+2+3...9)3 + 3*(1+2+3..9)2 + 2(1+2+3..9)

=453 + 3*452 + 2*45

So S/45  = 452 + 3*45 + 2

=2025 + 135 + 2

=2162

last digit is 2

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