For whatever reason (maybe I'm retarded) I'm just not following it.
While trick or treating in Clayton, NY (population 1,978), some friends play a counting game-they counted all the houses they visited while trick or treating (they all counted the same houses). Diego counted the houses by 5's, and had four houses left at the end. Madeline counted the houses by 7's, and had six houses left over at the end. Based on Diego's and Madeline's counts, how many houses did they visit? Is this the only possible answer? Explain why you are sure you found all possible counts.
this question is retarded, you'll have to guess how many people live in a house on average, then it is solvable
Brandon Smith
multiple answers. you are on the right track.
It boils down to a linear system of equations with 2 equations and 3 unknowns and you already got them.
The only limitation is that x, z, y are natural numbers.
The amount of houses can be 34,69,104,139... etc.
Tyler Powell
Yeah that's what I thought. I appreciate the confirmation and further guidance. This question has been killing me.
Brandon Wood
Tell me more please.
Benjamin Perez
So,
From what I got is X = (7Z+2)/5
That equals the two children's equations. I really have no idea how to pursue further like 7Z+2 part. How do I go about getting this broken down?
Adrian Garcia
X is a natural number, hence 7z+2 is divisable by 5. Which means it must end in 0 or 5. Which means z is a multiple of seven ending in 8 or 3.
Jaxson Wood
Ok you fucks:
X congruent 4 mod 5 => X = 5q + 4 X congruent 6 mod 7 => X = 7q + 6 Solve this linear system in integers for a solution that fits the population and you're golden
Jackson Ross
How does that not involve guesswork about how many people live in the house?
Leo Rivera
N mod 5 = 4 and N mod 7 = 6 The smallest N which satisfies both conditions is 34. Modularies will remain the same whenever N increments by 35 (5 x 7) So 69, 104, etc. are also possibilities. Assume every house has at least 1 occupant. 1975 is a multiple of 5 but that would mean at least 1979 occupants, which is impossible. So 1944 is the largest number of doorbells they could have rung.
Colton Stewart
Literally where in the question are they asking you about the population of each house? Point to it. I'll wait.
Nathan Morgan
>34,69,104,139... etc.
do you think two kids are going to cover more than 34 houses in one night?
Benjamin Bell
Doesn't appear explicitly.. But he asked "Explain why you are sure you found all possible counts." The population puts an upper limit on the number of houses in town if we assume each house has at least 1 occupant. Without that, the possibilities run into the billions.
Madeline and Diego are _really_ fast and _really_ gluttonous.
Justin Phillips
how the fuck should I know? we don't beg things from our neighbours at night once a year where I am from
Camden Wright
X congruent 4 mod 5 => X = 5q + 4 X congruent 6 mod 7 => X = 7s + 6 * 5q-7s=2 which is a linear diophantine equation and then solve for s and q which fit the problem, which is actually an overly complicated way of doing this
the easiest would be to just say X = 5q + 4, X is an integer on this line, what number for X makes sense
