divisibility - Finding out the last digit of number $3a/4$ knowing the last two digits of $a$

Natural number $a$ ends with digits $16$. If $a$ is not divisible by $8$, what is the last digit of $3a/4$?

Solution:

$$7$$

How do we arrive at that solution?

Answer

Note that $a = 100k + 16$ where $k$ is a positive integer.

Recall the divisibility criteria for $8$ : a number is divisible by $8$ if and only if its last three digits are.

Let $b$ be the last digit of $k$. Then, the last three digits of $a$ are $\overline{b16}$.

When is $\overline{b16}$ divisible by $8$? It is not difficult to see that this is a multiple of $8$ if and only if $b$ is a multiple of $2$, since $\overline{b16}$ is a multiple of $8$, if and only if $\overline{b00}$ is a multiple of $8$, if and only if $b$ is even.

In conclusion, $a = 100k + 16$ where the last digit of $k$ is odd. In other words, where $k$ is odd, since $a$ is not a multiple of $8$.

Now, $\frac a4 = 25k + 4$, and $\frac {3a}4 = 75k + 12$.

Note that $k$ is odd, so $75k$ is an odd multiple of $5$. Odd multiples of $5$ always end with $5$, so $75k + 12$ ends with $5+2 = 7$.