math problem help?

Question

In each round of a certain game for two people, the winner scores w points and the loser scores l points (w>l>0), ties are not possible, and all scores are integers. After several rounds, Bonnie leads Connie by a score of 30 to 25. If Connie has one exactly two rounds, compute w.

Easiest-to-understand answer gets the 10.

Answer 1

Dude above nearly got it right except he forgot the factor 'l' (loser scores 'l' points not '1' point):
Also, let us make life easy by substituting u = (x-2)

I found two different solution methods, plus the naive quadratic approach which is no good.

Connie's score is (2)(w) + (u)(l) = 25
2w + ul = 25 => u = (25-2w)/l
Bonnie's score is (u)w +(2)(l) = 30
uw + 2l = 30

[Connie] 2w + ul = 25
[Bonnie] uw + 2l = 30

It's easiest to solve by NOT forming the quadratic; instead we just try to pick integer solutions for u,w,l, noting that all three must be positive integers.
Note also that we must have w≥2 and l≥1, and u≥0
Also we can use the eqns to put upper bounds on u,w,l:
Noting ul = 25-2w and w≥2 => ul ≤ 21
Noting uw = 30-2l and l≥1 => uw ≤ 28 => u≤14

Also, since 2w + ul = 25, ul must be odd => u,l both odd

So we have very heavily constrained our solution space:
u odd, 0≤u≤14 → 1≤u≤13
l odd, 1≤l≤21
w≥2

[FIRST SOLUTION METHOD]

[C] 2w + ul = 25
[B] uw + 2l = 30

Multiply [C]*2 and [B]*u, so we get a "2ul" term which we will eliminate...
[C*2] 4w + 2ul = 50
[B*u] u²w + 2ul = 30u

Subtract [C*2] from [B*u]
[B*u - C*2] (u²-4)w = 30u-50 = 10(3u-5)
Call this eqn [D]:
[D] (u²-4)w = 10(3u-5)
and can determine l from [C']
[C'] l = (25-2w)/u

Now solve that for u odd, 1≤u≤13 and w≥2

u=1: (1²-4)w = 10(-2) ⇒ no integer solution for w
u=3: (3²-4)w = 10(4) ⇒ w=8, l=3,u=3 SOLUTION
u=5: (5²-4)w = 10(10) ⇒ no integer solution for w
u=7: (7²-4)w = 10(16) ⇒ no integer solution for w
u=9: (9²-4)w = 10(22) ⇒ no integer solution for w
u=11: (11²-4)w = 10(28) ⇒ no integer solution for w
u=13: (13²-4)w = 10(34) ⇒ no integer solution for w

***************************************************
Hence your UNIQUE SOLUTION is w=8, l=3,u=3⇒x=5
i.e. Bonnie won 3 of 5, Connie won 2 of 5, winner gets w=8 points, loser gets l=3 points
The question just asked to find w=8
***************************************************

Verifying it...
[C] 2*8 + 3*3 = 25
[B] 3*8 + 2*3 = 30


_____________________________

[ALTERNATIVE SOLUTION METHOD]

w(25-2w)/l + 2l = 30
2l² -30l + w(25-2w) = 0


The WRONG approach here is the quadratic approach, because it doesn't adequately exploit the knowledge that u,w,l are integer and small.

The RIGHT approach here is to look for integer factorizations of:
w(25-2w) = 2l (15-l)
Now note that these four brackets are all positive (NON-ZERO) and integers:
Remember also we know l odd, 1≤l≤21
and u odd, 1≤u≤13
Thus set l = 2L+1, u = 1+2U
=>
w(25-2w) = 2(2L+1) (15-(2L+1))
w(25-2w) = 2(2L+1) (14-2L)
w(25-2w) = 4(2L+1) (7-L)

So there must also be factor of 4 on the left.
It can't be in (25-2w) because that must be odd. Hence w must be divisible by 4. So set w=4W

(4W)(25-2(4W)) = 4(2L+1) (7-L)
W(25-8W) = (2L+1) (7-L)

Since all brackets must be positive including W(25-8W), that means W can only be 1 or 2 (W=0 not allowed)
W=1 gives 1(16) = (2L+1) (7-L) which has no solution since (2L+1) must be odd.
Hence W=2 (i.e. w=8) [SOLUTION!]
⇒ 2(9) = (2L+1) (7-L)
Since (2L+1) is odd, the 2 factor must divide into (7-L), hence (2L+1) must be 3 or 9.
Solution 2L+1=3, L=1
⇒ w=8, l=3,u=3⇒x=5
__________________________

And the WRONG approach is to try to algebraically solve the quadratic (what I initially tried):
2w² -25w + (30l-2l²) = 0
w = [ -(-25) ±√((-25)² - 4(2)(30l-2l²)) ] / 2(2)
= [ 25 ±√(625 - 240l + 16l²) ] / 4
Complelte the square for 16l² -240l...
= [ 25 ±√(16l² -240l + 30² - 275) ] / 4
= [ 25 ±√((4l -30)² - 275) ] / 4
Now let (4l -30) = m
w = [ 25 ±√(m² - 275) ] / 4
Now we have to solve by finding some m which both makes the sqrt integral, and makes w integral.
Clearly m is odd and >17...
so you just keep checking (17+2k)²-275 for squareness..
(17+2k)²-275 = 4k²+68k + (289-275) = 4k²+68k + 12
= 4(k²+17k + 3)
=> need only to check p(k) = (k²+17k + 3) for squareness
p(k) = k(k+17) +3
... p(3)=63 (nearly)... abandoned this approach.

Footnote: Plugging in our solution from above
w=8, l=3,u=3⇒x=5
into the quadratic:
2l² -30l + w(25-2w) = 0
w(25-2w) = 2l (15-l)
we get
8(25-16) = 6 (15-3)
8*9 = 6*12 = 72 correct

Answer 2

Let x = the number of rounds played
Connie has won 2 rounds
So Bonnie has won x-2 rounds
So Connie's score is 2w + x-2 = 25, or 2w+x = 27<-- Eq1
Bonnie's score is (x-2)w +2 = 30, or xw-2w= 28<-- Eq2
Solve Eq1 for x getting x = 27-2w
Substitute in Eq 2 getting:
(27-2w)(w) -2w =28
-2w^2 +25w -28= 0
Integer solution not possible