P&C q17

[Saumyojit]

a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one of ab, bc and ca is odd. abc is a multiple of 4 . The arithmetic mean of a and b is an integer and so is the arithmetic mean of a, b and c. How many such triplets are possible (unordered triplets)?


Now, product of two odd nos is always odd. So if a and b are odd (3,5,7,9) -->ab is odd.
abc is a multiple of 4 so c= 4 or 8

Then , How to proceed

 

[Dr.Peterson]

a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one of ab, bc and ca is odd. abc is a multiple of 4 . The arithmetic mean of a and b is an integer and so is the arithmetic mean of a, b and c. How many such triplets are possible (unordered triplets)?


Now, product of two odd nos is always odd. So if a and b are odd (3,5,7,9) -->ab is odd.
abc is a multiple of 4 so c= 4 or 8

Then , How to proceed

How many of a, b, c are odd?

How many choices are there for the even one(s)? (You answered that.)

What must be true of a and b for their mean to be an integer?

And so on. There are a lot of implications to think about before you do any actual combinatorics. Just keep thinking.

 

[Saumyojit]

How many of a, b, c are odd?

How many choices are there for the even one(s)? (You answered that.)

What must be true of a and b for their mean to be an integer?

How many of a, b, c are odd? Any two can be odd . 3 Combinations
Even can be any one out of 3 .

What must be true of a and b for their mean to be an integer? --> both has to be odd as both cannot be even as product has to be odd.

Then

 

[Dr.Peterson]

What must be true of a and b for their mean to be an integer? --> both has to be odd as both cannot be even as product has to be odd.

In other words, they must have the same parity (both odd or both even) because their sum must be even in order for the mean to be an integer; but the reason they can't both be even is that you've already determined that exactly one of the three is even.

So what do you now know about the three numbers?

And, does the fact that the mean of all three numbers is an integer add any further information?

Eventually, you'll have to think about what it means to say "How many such triplets are possible (unordered triplets)?" Since a and b are distinguished from c, how can you not distinguish orders?? Was that really stated in the problem, or did you add that?

 

[Saumyojit]

So what do you now know about the three numbers?

And, does the fact that the mean of all three numbers is an integer add any further information?

Obviously, the combination of three nos will be such so that it gives a integer divided by 3 which leads to (3,5,4);
(3,7,8) ; ( 5,9,4) ; (7,9,8)


But how to find these

Was that really stated in the problem

yes

 

[Dr.Peterson]

Obviously, the combination of three nos will be such so that it gives a integer divided by 3 which leads to (3,5,4);
(3,7,8) ; ( 5,9,4) ; (7,9,8)

But how to find these

I'd do what you probably did, more or less.

We know c is either 4 or 8, that a and b are odd, and that a+b+c is a multiple of 3. And a+b can be anything from 2+3=5 to 9+10=19.

If c=4=3+1, then a+b, which is even, must be 1 less than a multiple of 3, namely 8 or 14, which can be 3+5 or 5+9.

If c=8=9-1, then a+b, which is even, must be 1 more than a multiple of 3, namely 10 or 16, which can be 3+7 or 7+9.
So the possibilities are {3,5,4}, {5,9,4}, {3,7,8}, and {7,9,8}.

yes

Then in my mind, the problem is poorly stated. If you ignore order, then you can't distinguish between a, b, and c.