How many 4 digits numbers divisible by 5 can be formed with digits 0,1,2,3,4,5,6 and 6?

Question

How many 4 digits numbers divisible by 5 can be formed with digits 0,1,2,3,4,5,6 and 6

options:

a) $220$ b) $249$ c) $432$ d) $216$

MyApproach:

To form a 4 digit number divisible by 5 using given numbers

I make cases here:

Unit Digit is $0$ and other $3$ numbers can be formed in $7$ . $6$ . $5$=$210$

Unit Digit is $5$ and other $3$ numbers can be formed in $6$ . $6$ . $5$=$180$

Therefore,the required number is $390$

Is my approach right?Please correct me if I am wrong?

Answer 1

Your approach is fine except for you are missing the repeated digit. Here is a lazy solution:

Imagine we only have $0,1,2,3,4,5,6$, then your sums become:

• $0: 6\times5\times4=120$
• $5: 5\times5\times4=100$

Therefore there are at least $220$ and at most $390$ solutions, therefore the answer is b ($249$).

Answer 2

))U take two case for divisible by 5 case I last digit is 0 so last place. Fixed there 3 places can be filled in 6p3 ways then case 2 last place is 5 so 5 .5.4 solutions and more 29 solutions for at least 1 2 sixes