I'm not quite sure how Hannah or Guest got their answers, but I think there's a theorem called Chinese remainder theorem that you can use.
x = 5 (mod 9)
x = 3 (mod 7)
x = 1 (mod 5)
To use the chinese remainder therum, the gcd of all mods much be 1.
gcd(9, 7) = 1
gcd(7, 5) = 1
gcd(9, 5) = 1
It's a bit hard to explain, but there are good videos on it on yt. :))
One by "Randell Heyman."
Are mods are 9, 7, and 5.
(not 9) (not 7) (not 5)
x = (7*5) + (9*5) + (9*7)
First part (9 mod)
Now, it gets a bit confusing.
(9*5) and (9*7) will both be 0 in mod 9 (multiple of 9)
So we need to find a y*7*5 = 5 in mod 9.
y = 4.
Second part (7 mod)
x = (4*7*5) + (9*5) + (9*7)
(4*7*5) and (9*7) will both be 0 in mod 7.
(y*9*5) = 3 in mod 7.
y = 1, 45 = 3 in mod 7
Third part (5 mod)
x = (4*7*5) + (9*5) + (9*7)
(4*7*5) and (9*5) will both be 0 in mod 5.
(9*7*y) = 1 in mod 5
y = 2
Final equation
x = (4*7*5) + (9*5) + (9*7*2) = 311
Yayyy, it works. :))
I learned smth new today.
However, we need a 4 digit number.
311+(9*7*5)y = smallest 4 digit number
x = 1256
That has to be the most terrible explanation ever... but I tried. :)
=^._.^=
