Find the missing digit in a multiple of 48 The question I'm working on is as follows:
"In $62894\_52$, the hundreds place digit is missing. If $48 \mid 62894\_52$, find the missing digit."
I'm not really sure how to solve a problem like this. What I have so far is $48=3*2^4$, so $3\mid62894\_52$ and $2^4\mid62894\_52$. I don't know what to do next though. 
Can somebody give me a hint or point me to a Wikipedia page?
Thanks in advance!
 A: Here's a hint: since 16 divides 2000 evenly, $16 \mid 62894\_52$ is equivalent to $16 \mid \_52$.  First find out what hundredths digits will satisfy this; helpful is the fact that 16 doesn't divide 100 or 200 evenly, but $16 \mid 400$. 
Then, use the fact that $3 \mid 62894\_52$ exactly when the digits of $62894\_52$ add up to a multiple of three to eliminate all but one of your possibilities for the missing digit. 
A: $(62894052 + 100n) \mod 48  = 0 $
$(48(1310292+2n) + 36+4n) \mod48 = 0$
$(36+4n) \mod48 = 0$
From inspection, $4n = 12$, so $n=3$. 
A: Hint:
$3|x$ if the sum of $x$'s digits is also dividible by $3$. This nails down the list of possible digit values from 10 to a very manageable number.
Here, clearly $x$ is even, actually $4|x$, but $16|x$ imposes some more conditions on the missing digit.
A: It is just as easy to solve it for any multiple of $48$ with undetermined $100\,$'s digit $\color{#C00}j.$
Lemma $\ 48\:|\:10^4 i + 100 \color{#C00}{ j} + k\iff 4\:|\:k\ $ and $\: j \equiv -4i-k/4\pmod{\!12}$
Proof $\bmod 16\!:\ 10^4i+100j+k\equiv 4j+k,\:$ so $\:16\:|\:4j+k\!\iff\! 4\:|\:k,\:\ j\equiv -k/4\pmod{\! 4}$
Further, $ mod\ 3\!:\ 10^4i+100 j+k\equiv i+j+k,\:$ so $\:3\:|\:i+j+k\iff j\equiv -i-k\pmod{\!3}$
By CRT these two congruences for $\:j\:$ are equivalent to $\:j \equiv -4i-k/4\pmod{12}\ \ $ QED
In your case $\,62894\color{#C00}{\:\!j\:\!}52 = 6289\cdot 10^4 + 100 \color{#C00}j + 4052\:$ we have  $\:i = 6289,\ k = 4052,\:$ therefore applying the Lemma we deduce that $\bmod 12\!:\ j \equiv -4(6289)-4052/4\equiv -5 -4\equiv 3$.
A: The classic divisibility test for $2$-that of the last digit being even-extends to testing for $2^n$.  You check whether $2^n$ divides the last $n$ digits.  This works because $2$ divides $10$, so $2^n$ divides $10^n$, so all the digits in front of the $n$ don't matter to divisibility.
