What is the value of a + b + c?

Question

There are positive integers a, b, and c that solve this system of two equations:
c^2 - a^2 - b^2 = 101
a*b=72
What is the value of a + b+ c?

Answer 1

c² - a² - b² = 101
ab = 72

c² - a² - b² - 2ab = 101 - 2ab

ie c² - (a + b)² = 101 - 144 = -43

So (a + b)² - c² = 43

So [(a + b) + c][(a + b) - c] = 43

Now a, b and c are positive integers so a + b + c and a + b - c are positive integers

43 is prime and so 43 = 43 * 1 is the ONLY way it can be expressed as the product of two positive integers

Therefore

a + b + c = 43 and a + b - c = 1

Clearly therefore

c = 21 and a + b = 22

and since ab = 72

a = 18 or 4 and b = 4 or 18

Check

c² = 21² = 441

a² + b² = 4² + 18²
= 16 + 324
= 340

Thus c² - a² - b² = c² - (a² + b²)
=441 - 340
= 101

Answer 2

This is a tough one. There may be a better way to do this than mine but it works... and it helps if you have a spreadsheet program, otherwise you are going to be on a calculator for a long time.

using the 2 equations above, and substituting b = 72/a, you get:

c^2 = 101 + a^2 + (72/a)^2

All the possible values for a (being a positive integer) are:
1,2,4,6,8,9,12,18,36,72 (i.e. factors of 72)

this is where a spreadsheet helps:
plug in all the values for a in to the equation above and you get the following possible results for c^2:
246,281,441,1401,5286

Now since c must be a positive integer the only possibility that works is
c^2 = 441 which corresponds to a = 4 or 18
c = 21

all other values of c^2 are not integers.

finally
a + b + c = 4 + 18 + 21

= 43

if anyone finds a better way to do this... Please let me know too!


Addendum: I just saw Wal C's answer... Nice work! (better than mine I'm afraid... much less dependent on technology :) )

Answer 3

This can be done by brute force in a reasonable amount of time:

Given that a , b, and c are positive integers, a and b must be in one of these six pairs:
1, 72
2, 36
3, 24
4, 18
6, 12
8, 9

Substitute these pairs of integers in the first expression and solve for c.
The value of c that is a positive integer is the answer.
Then add up a, b, and c.

If you do this, you will find that 4 and 18 are a and b (or b and a, makes no difference). All the other pairs give non-integer values for c. This gives c=21.

So a+b+c=43.

Answer 4

For numeric values of a, b, and c, more information is needed. We have two equations with three unknowns. Since there are more unknowns than equations, we will not get just one solution for this. We can only get equations for each solution, giving us a value for each variable based on the others. The variables are dependant upon each other.

a*b=72

We can solve for a, giving us a = 72/b
So, given any real number value of b, we can find a.
If b=1, then a=72. If b=2, then a=35, and so on.

Plugging this back in to our other equation, we can solve for c:
c^2 - a^2 - b^2 = 101

c^2 – (72/b)^2 - b^2 = 101

c = [101 + (72/b)^2 + b^2]^(0.5)
Given any value of b, we can find c.
If b=0, then c=101. If b=1, then c=72.7…

Using the above equations for a and b, we can find a + b+ c in terms of b:
72/b+b+[101 + (72/b)^2 + b^2]^(0.5)

This can be solved in terms of a or c as well, I just chose b as an example.

Answer 5

c^2-a^2-b^2=101
a*b=72
There are only 12 positive integers that satisfy the equation, a*b=72 for the values of a&b:
1&72, 2&36, 3&24, 4&18, 6&12, 8&9
and none of these combinations will satisfy:
c^2-a^2-b^2=101if C is also a positive integer.
Go figure?

Answer 6

c^2 = 101 + a^2 + b^2 = 101 + (a + b)^2 - 2ab = (a+b)^2 - 43
=> (a+b)^2 - c^2 = 43 = (a + b + c) (a+b - c) = 43

well, I need one more thing to finish the answer

Answer 7

James Chan is correct.

43 is a prime number. Therefore, 43=43*1.
=>a+b+c=43, a+b-c=1

Answer 8

43.

a or b can either be 4 or 18, and c is 21.

Answer 9

Undefined. You have 3 unknowns & only 2 equations.

Answer 10

in terms of which variable?