#1**+1 **

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)

RiemannIntegralzzz Apr 6, 2021

edited by
RiemannIntegralzzz
Apr 6, 2021

edited by RiemannIntegralzzz Apr 6, 2021

edited by RiemannIntegralzzz Apr 6, 2021

edited by RiemannIntegralzzz Apr 6, 2021

edited by RiemannIntegralzzz Apr 6, 2021

edited by RiemannIntegralzzz Apr 6, 2021

edited by RiemannIntegralzzz Apr 6, 2021

edited by RiemannIntegralzzz Apr 6, 2021

edited by RiemannIntegralzzz Apr 6, 2021

#2**+2 **

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

Jamesaiden
Apr 6, 2021

#4**+3 **

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

Melody Apr 6, 2021

#5**+3 **

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! :)

RiemannIntegralzzz
Apr 6, 2021