# What will be probability in this case?

A five digit number is formed without repetition of the digits $1,2,3,4,5$ in a random order. What will be the probability that the number is divisible by $4$?

I tried to solve and after a bit of calculations and observations, I came to know that the number will be divisible by $4$ if the last 2 digits of the number are $12, 24, 32$ or $52$. But it took me so much time to observe this trend.

So my answer is as follows:

Number of favorable cases $= 3! \times 4 = 24$

Number of exhaustive cases $= 5! = 120$

So the probability $= 1/5$

My question is that, is there any other method of solving it quickly?

• That seems like the best way to solve it. – T.J. Gaffney Jan 8 '17 at 1:39

The "quick" way to see it is to write the number $a_5a_4a_3a_2a_1$ (the digits) in its base $10$ representation: $$a_5a_4a_3a_2a_1 = a_1+10a_2+10^2a_3+10^3a_4+10^4a_5$$ Now, a number $b$ is divisible by $4$ if $b\pmod{4} = 0$. Lets look at this number mod $4$: $$a_1\pmod{4}+2a_2\pmod{4}+0+0+0\pmod{4} = 0$$ So, we need to have that: $$a_1 = 2a_2\pmod{4}$$ If you look at the cases you have, they're precisely when this is true.
• That denotes "modular arithmetic" (also known as "clock arithmetic"). Essentially, when we tell time, what happens after $12$ o clock? It becomes $1$. In this way, the numbers for time kind of "wrap around", so if I start at 2 pm, and it's been 37 hours, it isn't 39 pm. Instead, we say that 2pm is 14 hours into the day, add 37 hours to get 51 hours, then remove 24 hours once to get 27, then again to get 3, so it's 3 am two days from now. To look at a number $\pmod{4}$, what we do is we say "If I subtract off enough $4$'s, this will be between $0$ and $3$. It turns out this can tell us – Mark Jan 8 '17 at 1:48
• A lot about the number. A few things are that $(a+b)\pmod{r} = a\pmod{r}+b\pmod{r}$, and $ab\pmod{r} = (a\pmod{r})(b\pmod{r})$, as $a\pmod{r}$ is generally much smaller than $a$, it can simplify arithimetic a lot. Additionally, it can tell us a ton about a number. As an example, if a number $x$ is zero mod any other number $y$, then $x$ is a multiple of $y$. So, $52\pmod{4} = 0$ (try subtracting out $4$'s to see this), so we know that $52 = 4\times k$ for some $k$ (in this case, $k = 13$). – Mark Jan 8 '17 at 1:50
Hint: A number is divisible by $4$ if its last two digits are divisible by $4$.