Multiplication Magic
Problem
Here is something to try on your friends.
Ask a friend for a three digit number and suppose she says 314 then you immediately echo her number in your reply "Did you know that 241314 is exactly divisible by 37?"
Another friend gives 628 and you say "Amazing, 628371 is exactly divisible by 37 as well!"
When they check they find you are correct.
In general if the friend gives the number $abc$ you give $abcdef$ or $defabc$ where $a+d = b+e = c+f = x$ where $1 \leq x \leq 9$.
Why does this work?
You can even do the same trick with nine digit numbers. Your friend suggests 143 and you say, immediately "Another multiple of 37 is 143110635!"
You use exactly the same technique. Take your friend's number and think of two more three digit numbers such that the sum of the first digits, the sum of the second digits and the sum of the third digits are all the same.
Explain why the trick works for nine digit numbers.
Getting Started
111, 222, 333,... etc are all divisible by 37.
A second hint is to use 1000= 999+1 when breaking the number into blocks of 3 digits
Student Solutions
This problem was solved by Alan Riddell of Madras College, Scotland; Justin Sinz, age 16, Skyview HS, Billings, MT, USA; Joel Tay, age 13 ACS, Singapore, and Ling Xiang Ning, Tao Nan School, Singapore.
This is Alan Riddell's solution:
Let the number $n$ be written using the six digits $abcdef$ where
$a + d = b + e = c + f = x.$
$$\begin{eqnarray} n &=& 10^5a+ 10^4b + 10^3c + 10^2d + 10e
+ f\\ &=& 99900a + 9990b + 999c + 100(a + d) + 10(b + e) +
(c + f)\\ &=& 999 (100a + 10b + c) + 111x\\ &=&
37[27(100a + 10b + c) + 3x] \end{eqnarray}$$ and so 37 is a factor
of $n$.
Also if $n$ has nine digits $abcdefghi$ where $a + d + g= b + e + h
= c + f + i= x$,
$$\begin{eqnarray} n &=& 10^8a + 10^7b + 10^6c + 10^5d +
10^4e + 10^3f + 10^2g + 10h +i.\\ &=& 99999900a + 9999990b
+ 999999c + 999(100d + 10e + f) + 100(a + d+ g) + 10(b + e + h) +
(c + f + i)\\ &=& 999999 (100a + 10b + c) + 999(100d + 10e
+ f) + 111x\\ &=& 37[27027(100a + 10b + c) + 27(100d + 10e
+ f) + 3x] \end{eqnarray}$$ and so 37 is a factor of $n$.
This process can be applied to 12 digit numbers, to 15 digit
numbers and to any numbers where the number of digits is a multiple
of 3.
Justin Sinz used a slightly different method as follows: Let the
number be represented by $abcdef$ where:
$$abcdef=10^5a + 10^4b + 10^3c + 10^2d + 10e + f.$$
If an integer $p$ gives a remainder of $q$ when divided by $n$, we
say that '$p$ is congruent to $q$ (mod $n$)' which is written $p
\equiv q$ (mod $n$). From this, we can see that if $p$ and $q$
satisfy $p - q \equiv 0$ (mod $n$), then $(p - q)$ is divisible by
$n$. Just for fun, let's write out
$$\begin{eqnarray} 10^5 &\equiv& g (\bmod 37)\\ 10^4
&\equiv& h (\bmod 37)\\ 10^3 &\equiv& i (\bmod
37)\\ 10^2 &\equiv& j (\bmod 37)\\ 10 &\equiv& k
(\bmod 37)\\ 1 &\equiv& l (\bmod 37) \end{eqnarray}$$
where $g,h,i,j,k,l$ are constants to be determined. It is found
that $g=j=26, h=k=10$, and $i=l=1$. Now in the top equation, add
and subtract $g,h,i,j,k,l$ from their corresponding powers of ten;
for example, replace $10^5a$ with $a(10^5 - g + g)$ (with the
exception of $f$). Regroup so that, for example,$a(10^5 - g + g)$
becomes $a(10^5 - g) + ag$, and so on. Now gather together the
terms that look like $a(10^5 - g)$ onto the left side of the right
side of the equation, and terms like $ag$ to the right side of the
right side of the equation to give:
$$abcdef = [a(10^5 - g) + ... + e(10 - k)] + ag + ... fl.$$
The term in the brackets is clearly always divisible by 37;
therefore $abcdef$ is divisible by 37 if $ag + bh + ci + dj + ek
+fl$ is divisible by 37. But $g=j=26, h=k=10,$ and $i=l=1$;
therefore
$$ag + bh + ci + dj + ek +fl = 26(a + d) + 10(b + e) + (c +
f).$$
This is always divisible by 37 if $a + d = b + e = c + f = x$,
because the expression then equals $37x$. Hence the first theorem.
In the second theorem, let the number be $abcdefghi$. One can use
exactly the same technique, noting that $10^8 \equiv 26$ (mod 37),
$10^7 \equiv 10$ (mod 37), and $10^6 \equiv 1$ (mod 37).