N is a four-digit positive integer. Dividing N by 9, the remainder is 5. Dividing N by 7, the remainder is 4. Dividing N by 5, the remainder is 1. What is the smallest possible value of N?
The smallest 4-digit n =1256
1256 mod 9 =5
1256 mod 7 =3
1256 mod 5 =1
How did you arrive at this answer Hannah?
I do not understand your working either guest. What is CRT + MMI ?
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. :)
=^._.^=
Read the question carefully!
N mod 9 ==5
N mod 7 ==4 [NOT 3]
N mod 5 ==1
See this link here: https://web2.0calc.com/questions/congruences
It doesn't matter that you copied the question incorrectly catmg. It is not YOUR homework.
You explained your method the best you could and that is all that counts.
The asker is always expected to check then accept, edit or reject the answers given to them.
I learned from your answer in that I looked up the Chinese remainder theorem. Thanks
This is the video that I learned from. I found it excellent.
It is from a channel called "Maths with Jay"
I got the answer 1166 (which is the same as our first guest's answer.)