# 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!

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What is $|$ being used for here? –  David Sep 20 '12 at 22:29
To show that $48$ divides 62894_52. For example $10/2=5$ so $2\mid10$ and $5\mid10$. –  Noah Klein Sep 20 '12 at 22:31
The | means divides. –  Amzoti Sep 20 '12 at 22:32

## 5 Answers

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.

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Really helpful. Thanks so much! –  Noah Klein Sep 21 '12 at 0:37
You're welcome! –  JakeR Sep 21 '12 at 0:39

It is just as easy to solve it for any multiple of $48$ with undetermined $100\,$'s digit $\rm\:\color{#C00}j.$

Lemma $\rm\ 48\:|\:10^4 i + 100 \color{#C00}{\ j} + k\iff 4\:|\:k\$ and $\rm\: j \equiv -4i-k/4\pmod{12}$

Proof $\rm\ mod\ 16\!:\ 10^4i+100\ j+k\equiv 4j+k,\:$ so $\rm\:16\:|\:4j+k\iff\ 4\:|\:k,\:\ j\equiv -k/4\pmod 4$

Further, $\rm mod\ 3\!:\ 10^4i+100\ j+k\equiv i+j+k,\:$ so $\rm\:3\:|\:i+j+k\iff j\equiv -i-k\pmod{3}$

By CRT these two congruences for $\rm\:j\:$ are equivalent to $\rm\:j \equiv -4i-k/4\pmod{12}\ \$ QED

In your case $\rm\:62894\color{#C00}{\,j\,}52 = 6289\cdot 10^4 + 100\ \color{#C00}j + 4052\:$ we have $\rm\:i = 6289,\ k = 4052,\:$ therefore applying the Lemma we deduce that, $\rm\:mod\ 12\!:\ j \equiv -4(6289)-4052/4\equiv -5 -4\equiv 3.$ Finally, being a decimal digit, $\rm\:0\le j\le 9\:$ hence $\rm\:j \equiv 3 + 12\,n\:\Rightarrow\:j = 3.$

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$(62894052 + 100n) \mod 48 = 0$

$(48(1310292+2n) + 36+4n) \mod48 = 0$

$(36+4n) \mod48 = 0$

From inspection, $4n = 12$, so $n=3$.

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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.

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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.

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