Maths
Maths Puzzle
Numerical Ability
Permutation and Combination
How many 3 digit numbers can be formed using digit 1 to 6,without repetition such that number is divisible by the digit at its unit place ??
Read Solution (Total 2)
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- in this case
_ _ 1 first the last digit is 1 then 5*4=20 cases can be possible
then
_ _ 2 last digit is 2 then 5*4 that is 20 cases are possible
then
_ _ 3 last digit is three then for the divison of the number by three sum of all the numbers should be multple of three
then these cases can be possible that are 12,15,21,24,42,51 that are total 6 cases
then for last digit is four then these cases can be possible that are
12,16,26,32,36,56,52 and 62 total 8 cases
for last digit to 5 5*4 cases can be possible
last digit is 6 then 6 cases that are 12,15,21,24,42 and 51 are possible
total of 80 cases can be possible - 10 years agoHelpfull: Yes(1) No(2)
- Number with unit digit 1,2,3,5 and 6 and divisible by the same will be 5*5! i.e 600 ways
now we are left with unit digit 4 and divisible by 4 case so in this case last 2 digits can only be 24 and 64 so 2*4! i.e 48 ways
so total = 600+48 = 648 , it should be i believe - 7 years agoHelpfull: Yes(0) No(0)
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