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Question Number 52379 by Necxx last updated on 07/Jan/19

The sum of the digits in the unit place of all four digit numbers formed using the numbers 3 4 5 andL 6 without repetition is a)432 b)108 c)36 d)18

$${The}\:{sum}\:{of}\:{the}\:{digits}\:{in}\:{the}\:{unit} \\ $$$${place}\:{of}\:{all}\:{four}\:{digit}\:{numbers} \\ $$$${formed}\:{using}\:{the}\:{numbers}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:{and}\mathscr{L} \\ $$$$\mathrm{6}\:{without}\:{repetition}\:{is} \\ $$$$\left.{a}\left.\right)\left.\mathrm{4}\left.\mathrm{32}\:{b}\right)\mathrm{108}\:{c}\right)\mathrm{36}\:{d}\right)\mathrm{18} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jan/19

4our digit numbers=4!=24 while adding we have to add each digit(3,4,5,6) 6 times so sum of digut in unit place is 6(3+4+5+6)=108

$$\mathrm{4}{our}\:{digit}\:{numbers}=\mathrm{4}!=\mathrm{24} \\ $$$${while}\:{adding}\:{we}\:{have}\:{to}\:{add}\:{each}\:{digit}\left(\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\right) \\ $$$$\mathrm{6}\:{times} \\ $$$${so}\:{sum}\:{of}\:{digut}\:{in}\:{unit}\:{place}\:{is} \\ $$$$\mathrm{6}\left(\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}\right)=\mathrm{108} \\ $$

Commented by Necxx last updated on 07/Jan/19

sorry for my funny questions but why did you multiply by 6 times. When I solved I had 18 by just adding all the digits but your answer is the correct one.Thanks in advance.

$${sorry}\:{for}\:{my}\:{funny}\:{questions}\:{but} \\ $$$${why}\:{did}\:{you}\:{multiply}\:{by}\:\mathrm{6}\:{times}. \\ $$$${When}\:{I}\:{solved}\:{I}\:{had}\:\mathrm{18}\:{by}\:{just} \\ $$$${adding}\:{all}\:{the}\:{digits}\:{but}\:{your} \\ $$$${answer}\:{is}\:{the}\:{correct}\:{one}.{Thanks} \\ $$$${in}\:{advance}. \\ $$

Commented by mr W last updated on 07/Jan/19

there are 3!=6 numbers whose last digit is 3, 4, 5, 6 respectively.

$${there}\:{are}\:\mathrm{3}!=\mathrm{6}\:{numbers}\:{whose}\:{last} \\ $$$${digit}\:{is}\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6}\:{respectively}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 08/Jan/19

− − − −^3 unit place 3→(3×2×1)=6nos − − − −^4 unit place 4→(3×2×1)=6 nos − − − −^5 unit place 5 →(3×2×1)=6nos − − − −^6 unit place 6 →(3×2×1)=6 nos now add them....you have to add=6(3+4+5+6)=108..

$$−\:−\:−\:\overset{\mathrm{3}} {−}\:{unit}\:{place}\:\mathrm{3}\rightarrow\left(\mathrm{3}×\mathrm{2}×\mathrm{1}\right)=\mathrm{6}{nos} \\ $$$$−\:−\:−\:\overset{\mathrm{4}} {−}\:{unit}\:{place}\:\mathrm{4}\rightarrow\left(\mathrm{3}×\mathrm{2}×\mathrm{1}\right)=\mathrm{6}\:{nos} \\ $$$$−\:−\:−\:\overset{\mathrm{5}} {−}\:{unit}\:{place}\:\mathrm{5}\:\rightarrow\left(\mathrm{3}×\mathrm{2}×\mathrm{1}\right)=\mathrm{6}{nos}\: \\ $$$$−\:−\:−\:\overset{\mathrm{6}} {−}\:{unit}\:{place}\:\mathrm{6}\:\rightarrow\left(\mathrm{3}×\mathrm{2}×\mathrm{1}\right)=\mathrm{6}\:{nos} \\ $$$${now}\:{add}\:{them}....{you}\:{have}\:{to}\:{add}=\mathrm{6}\left(\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}\right)=\mathrm{108}.. \\ $$

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