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(#M40033140) ELITMUS Numerical Ability question permutaion and combination Keep an EYE

How many six digits nos. can be formed using digits 1 to 6 such that no. is always divisible by digit at its unit place??

Asked In Elitmus falak khan (11 years ago)

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RAKESH  (11 years ago)

unit digit=1,no. is always divisible by 1 => 5!

unit digit=2, no. is always divisible by 2 => 5!

unit digit=3, no. is always divisible by 3 => 5! [sum of digits=1+2+3+4+5+6=21,always divisible by 3]

unit digit=4, no. will be divisible by 4 if 10th digit 2 or 6 => 2* 4!

unit digit=5, no. is always divisible by 5 => 5!

unit digit=6, no. is always divisible by 6 => 5! [as, no. ending with 6 is divisible by 2 & sum of digits=21 so divisible by 3 , hence divisible by 6]

total= 5! + 5! + 5! + 2*4! +5! + 5! = 5*5! + 2*4! = 648

Is this solution Helpfull? Yes (22) | No   

Avinash  (11 years ago)

hello manikandan, why are you simply copying above answers... rather post your own better explanation... I saw your answers in other posts also... so telling this

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Mani kandan  (11 years ago)

_ _ _ _ _ 1 =>5!=5*4*3*2*1=120
_ _ _ _ _ 2 =>5!=5*4*3*2*1=120
_ _ _ _ _ 3 =>5!=5*4*3*2*1=120
_ _ _ _ _ 4 =>2*4*3*2*1=48 (Tenth digit can have either 2 or 6)
_ _ _ _ _ 5 =>5!=5*4*3*2*1=120
_ _ _ _ _ 6 =>5!=5*4*3*2*1=120
Total: 120*5 + 48 =648 ans

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