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


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


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?


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 ($249$).

  • 2
    $\begingroup$ For the less lazy solution and this is treated as a multiset with two sixes available, Case 1: at most one six occurs, solved above as $220$. Case 2: two sixes occur. Case 2a: the last digit is a zero, pick which space is not used by a six and pick which digit it is, $3\cdot 5$ choices. Case 2b: the last digit is a five. Case 2bi) the extra digit is a zero, can't be first space so $2$ choices. Case 2bii) the extra digit is not a zero, $3\cdot 4$ choices, for total of $220+15+12+2=249$ choices. $\endgroup$ – JMoravitz Oct 10 '15 at 16:25

))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


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