+486 Sorry, it is not rendering for some reason.
Here is the question without Latex:
How many integers n satisfy 0 is less than n is less than 60 and 4n=2 (mod 6)
+150 0 < n < 60
\(4n\equiv 2\pmod 6\)
\(n\equiv \frac{2}{4}\pmod6\)
\(n\equiv \frac{2+6}{4}\equiv2\pmod6\)
n = 6k + 2
k = 0, 1, 2, ..., 9
total : 10
+118687 My attempt
All pronumerals are integers
LaTex not displaying so here is a pic that I took of it.
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\(0
which means there are 18 possible vaues of g and hence 18 possible values of n
LaTex:
0 4n=2\mod6\\
2n=1\mod3\\
2n=3k+1\\
2n-3k=1\\
\text{One solution is n=2, k=1 }\\
2(2\qquad)-3(1\qquad)=1\\
2(2+3g)-3(1+2g)=1\\
so\\
n=2+3g\\
0<2+3g<60\\
-2<3g<58\\
-0.6 0\le g\le17
+486 Sorry @Melody and @Jamesaiden, they were both wrong. The correct answer was 20. Here was the solution they gave:
The residue of 4n(mod 6) is determined by the residue of n (mod 6). We can build a table showing the possibilities(it doesn't render, but here it is with words)
n (mod 6) 0 1 2 3 4 5
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4n (mod 6) 0 4 2 0 4 2
As the table shows, 4n=2 (mod 6) is true when n=2 or n=5 (mod 6). Otherwise, it's false.
So, our problem is to count all n between 0 and 60 that leave a remainder of 2 or 5 (mod 6). These integers are 2, 5, 8, 11, 14, 17, ... , 56, 59.
There are $\boxed{20}$ integers in this list.
Thank you both for trying, however! :)
