CPhill

avatar
Имя пользователяCPhill
Гол130557
Membership
Stats
Вопросов 56
ответы 43469

 #5
avatar+130557 
0

However....I'm sure  that the answer isn't 450, either....

 

Assuming that we can't begin with 0, there are 900 integers from 100 - 999 and half of these will be even = 450....and some of those will have repeated digits.....so.....the answer is less than 450

 

Here's a "brute force"  method

 

Let's look at all the integers that end in 0

 

There are  10 of these in each group of 100.....and we have 9 groups of 100 from 100-999

So....that's 10 * 9  = 90 possible integers.....but...we have to discard 10 of these (100, 200, 300. etc.) because of the repeated 0....so......that makes 80 possible numbers ending in 0

 

Now.....let's look at all the numbers ending in 2

Again there are 10 of these in each group of 100.....so, again, that's 90 possible numbers

But.....from these we must subtract all the numbers that have 2 as a middle digit and 2 as an ending digit.....and there are 9 of these   (122, 222, 322, 422, etc )...so that gives us 81 possible numbers....

Also.....we have to subtract the numbers that start with 2 and have 2 as an ending digit....there are 9 of these, as well  (202, 212, 232, 242, etc. ).....(note that we have already counted '222")....so.....the possible numbers ending in 2 with no repeated digits  = 90 - 9 - 9  =  72

 

And the other numbers ending in 4,6 or 8  will similarly occur 72 tmes each

 

So.....the total combinations ending in an even number with no repeated digits  = 80 +  4 * 72  =

 

80 + 288  =   368

 

As Melody always says, "That's what I think!!! "

 

 

cool cool cool

14 июн. 2017 г.