Menu Close

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-




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

Leave a Reply

Your email address will not be published. Required fields are marked *