indiheats wrote:
Hi Banuel,
I get hung up on when i create the equation 15x + 18y = 38700.... When I see a equation like this, should i automatically assume that mutiple combinations of x and y are possible to satisfy the equation or are there instances where I should actually work out the math.... I spend a lot of time contemplating this, although I see the obvious answer in C of two liner equations...
thoughts?
Thanks
Your thought process should be how can we make these numbers more manageable. 15 and 18 both share 3 as a factor, but 38700 looks pretty gnarly. A quick check confirms that 3 is a factor, 3+8+7=18, which is divisible by 3
Now we have the equation in something easier to work with 5x + 6y = 12,900.
It still looks pretty daunting. So here's my thought process, what two values when added give us 12,900, in other words, we're asking what gives us 12,000 + 900
So that equation now becomes, 5x+6y = 12,000 + 900
Can we get an x such that we get 12,000 or 900. Yes.
Can we get a y such that we can get 12,000 or 900. Yes. 120 is divisible by 6, and 90 is divisible by 6.
What does that mean for us?
Well, we can have a case x=2,400 and G = 150
5*2400 + 6*150 = 12,900
OR
We can have a case where x=180 and G= 2000
5*180 + 6*2000 = 12,900
You don't have to actually do the arithmetic. Just do a quick sense check, can we have multiple values for x and y, such that we can get 12000 + 900. Use divisibility rules, if x can give us either 900 or 12000 when multiplied by 5, both are divisible by 5, and if Y can give us either 12000 or 900 when multiplied by 6. Both are divisible by 6. So we can get different values for x and y, and still satisfy 12,900.
Hope that helped someone, I know this post is a bit dated.
- a quick look at the equation will tell you that 12900 is a very large number compared to 5 and 6. Hence it is quite likely to fit in multiple combinations of 5 and 6. INSUF
- let x=0, 12900 is divisible by 6(digits of 12900 add up to 3 and is an even number) so y will be an integer. This will give us one combination. Now let y=0, 12900 is obviously divisible by 5--> second combination. INSUF