Question 1169499: how many three digit numbers can be formed from the set 0-9 if repetition are allowed and the number must be a).even b).divisible by 5 c).divisible by 10
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! how many three digit numbers can be formed from the set 0-9 if repetition are allowed and the number must be
a)even
--> the 1's digit must be 0, 2, 4, 6 or 8
Units is 1 of 5
Tens is 1 of 10
Hundreds is 1 of 10
---> 5*10*10
========================
b)divisible by 5
---> Units is 5 or 0
Tens is 1 of 10
Hundreds is 1 of 10
---> 2*10*10
======================
c)divisible by 10
---> Units is 0
Tens is 1 of 10
Hundreds is 1 of 10
---> 10*10
===============================
All assuming leading zero is acceptable, eg, 012.
If not, you need to specify that.
Answer by ikleyn(52864)  (Show Source):
You can put this solution on YOUR website! .
In such problems, the term "three digit number" means, by default, the number having NON-ZERO leading digit.
Therefore,
(a) the last, ones digit, is any of 5 options 0, 2, 4, 6, 8.
the tens digit is any from 0 to 9 : 10 options;
the leading digit: any from 1 to 9 : 9 options.
Total amount of such numbers is 9*10*5 = 450.
The same answer you can obtain by different way
noticing that there are 900 three-digit numbers from 100 to 999,
and exactly half of them are even numbers.
(b) If a three-digit number is divisible by 5, its ones digit must be 0 or 5.
So, the amount of such three-difit numbers is 9*10*2 = 180.
Again, it is exactly  part of the total 900 three-digit numbers.
(c) Let me be short in this last case:
The amount of the three-digit numbers divisible by 10 is exactly  part
of the total 900 three-digit numbers, i.e. 90, in total.
Solved.
All questions are answered.
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