Nat is thinking of a four-digit counting number. The thousands digit and the hundreds digit are the same. The sum of all 4 digits is 28. The four-digit number is an odd multiple of 5. Exactly two digits are even. What is Nat's number?

The number is 8875

Step-by-step explanation:

Let the numbers be ABCD

Now, A has only 8 options because it's number can be from 1 to 9 excluding 5 because it is reserved for the last number D from the question.

Since A and B are to be the same from the question, then B also has 8 options.

C has only 7 options from 1 to 9 (I didn't say from 0 to 9 for a reason you'll soon see) excluding 2 numbers designated for A and B, then for D.

And D has only one option which is 5.

Since exactly two must be even, then the first two numbers have to even because only the first 3 numbers have possiblity of being even and if I pick a number an odd number for the first number, the second must be the same. Therefore, there is no way exactly two can be even except that the first two are even and for this reason C cannot not be 0 like I said earlier on.

If we pick 6 for the first two, it means we have 66C5. Now since adding the numbers gives 28, there is no single digit that can fit C. The case is the same if we pick 4 for the first and second number since 6 is larger than 4. Hence the only option is picking 8 for the two numbers, so we have

88C5

If C is 7, addition of the digits give 28. Therefore our number is

8875

This satisfies all the condition stated in the question.

8875

Step-by-step explanation:

So first we can have 4 empty blank spots like this:              

Then we can say how would you get an odd multiple of 5? 5 is divisible only for numbers that end with 5 and 0, so since 0 is even and 5 is odd, we will pick 5 to be the last digit. Now we know that the thousands and hundreds digits are the same. The problem also says that there are exactly 2 even digits. So we will assume that the thousands and hundreds digit are even because if we made the tens digit even then made one of the remaining digits even we would have to add a 3rd even digit because the first two digits are the same. Since the sum is 28, to narrow it down, I'm going to do 28-5=23. So now I know that the thousands, hundreds, and tens digit have to equal 23 when they're added together. I am doing some trial-and-error here and I will start with the highest even number since our sum is so big. 8+8=16. So now I am going to narrow it down even more. 23-16=7. Since 7 is odd we know have our four-digit number.

8875

the two may have that, even a breakdown is probably offered