Question #23071

how many three digit numbers can be formed from the set 0,1,2,...,9 if repetitions are not allowed and the number must be divisible by a)10 b)25.

Expert's answer

a) is divided by 10

There can be:

& 1) only one digit on the third place = 0, because the number must devisible by 10 ;

& 2) 9 digits on the first place (all digits from the set except 0),

& 3)8 digits on the 2-nd place (all 10 digits except two we have already used for the first and third places)

So the total number of possibilities is (by multiplying rule) N = 1*9*8 = 72

b) is divided by 25

The number is divided by 25 if and only if last two digits are divided by 25.

There are 3 variances for the last two digits

& 1) 25 or 75. In that case we can put 7 different digits on the first place (all 10 except 0, 5, 7 or 2)

& 2) 00. In that case we can put 9 different digits on the second place (all ten except 0)

So the total number of possibilities is (due to addition and multiplication rules):

& N = 2*7+ 1*9 = 23

There can be:

& 1) only one digit on the third place = 0, because the number must devisible by 10 ;

& 2) 9 digits on the first place (all digits from the set except 0),

& 3)8 digits on the 2-nd place (all 10 digits except two we have already used for the first and third places)

So the total number of possibilities is (by multiplying rule) N = 1*9*8 = 72

b) is divided by 25

The number is divided by 25 if and only if last two digits are divided by 25.

There are 3 variances for the last two digits

& 1) 25 or 75. In that case we can put 7 different digits on the first place (all 10 except 0, 5, 7 or 2)

& 2) 00. In that case we can put 9 different digits on the second place (all ten except 0)

So the total number of possibilities is (due to addition and multiplication rules):

& N = 2*7+ 1*9 = 23

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