Question

Suppose that a and b are two positive intgers with the same number of digits for which their average is obtained by wriing them side by sidend placing a decimal point between them, to form the number b.a For any two number a and b, with a >b, it must be true that :
a) a-b=1 b) a/b = 5/4 c) a/b=/5 d) a+b=9
e) a-b=50

Answer 1

Since you are taking the average of two numbers, you are essentially doing:

\[Average=\frac{ a+b }{ 2 }\]
So you will always divide by 2. Now a+b is an integer. So if you divide an integer by 2 you can get either another integer (10/2 = 5) or an integer and a half (11/2 = 5.5).

Since the resultant number takes the form b.a , that means that a must have a leading digit of 5, since you can't really have a leading digit of 0 (and the only two possible digits after the decimal are 0 and 5).

So a=5,50,500, etc.

Remembering that a>b, if you just try out some numbers for b with a given a you will realize a pattern:
| a | b | Avg. | b.a
5 1 3 1.5
5 2 3.5 2.5
5 3 4 3.5
5 4 4.5 4.5
50 32 41 32.5
50 47 48.5 47.5
50 49 49.5 49.5

As you can see from the table above, for the conditions to be met b must be one less than a, or a-b=1.

Hope that helps!

Answer 2

Oh I see. Thank you so much, your explanation is very clear and easy to follow! Thanks! :)

Answer 3

You're welcome! :D