Digit puzzles

Mathematicians have always loved playful puzzles where the only rules are arithmetic. Many of these puzzles look mysterious until you peek behind the curtain , then they become beautiful patterns. Some are over $4000$ years old.

Concept

Magic squares. A magic square is a square grid filled with numbers so that every row, every column, and both diagonals add to the same total , the magic sum. The most famous is the $3 \times 3$ Lo-Shu square:

$\begin{array}{|c|c|c|}\hline 2 & 7 & 6 \\ \hline 9 & 5 & 1 \\ \hline 4 & 3 & 8 \\ \hline\end{array}$

Every row, column, and diagonal sums to $15$. The numbers $1$-$9$ each appear exactly once. A medieval Indian magic square attributed to Khajuraho temples uses the numbers $1$-$16$ on a $4 \times 4$ grid with magic sum $34$.

General formula. A magic square using $1, 2, \dots, n^2$ has magic sum $\dfrac{n(n^2+1)}{2}$. For $n = 3$ this gives $\dfrac{3 \cdot 10}{2} = 15$. For $n = 4$, $\dfrac{4 \cdot 17}{2} = 34$. For $n = 5$, $65$.

Palindromic numbers. A palindrome reads the same forwards and backwards: $121, 9999, 12321$. Multiplying small palindromes can give larger palindromes: $11 \times 11 = 121$, $11 \times 11 \times 11 = 1331$, $11 \times 11 \times 11 \times 11 = 14641$. Up to $11^4$ the pattern (rows of Pascal's triangle) gives palindromes.

Cyclic numbers. The decimal expansion of $\tfrac{1}{7} = 0.\overline{142857}$ has a remarkable property:

$1 \times 142857 = 142857, \quad 2 \times 142857 = 285714, \quad 3 \times 142857 = 428571,$ $4 \times 142857 = 571428, \quad 5 \times 142857 = 714285, \quad 6 \times 142857 = 857142.$

Every product is a cyclic rotation of the digits $1, 4, 2, 8, 5, 7$. And $7 \times 142857 = 999999$.

Digit-sum tricks. Pick any number, say $173$. Reverse: $371$. Subtract the smaller from the larger: $371 - 173 = 198$. Now $1 + 9 + 8 = 18$, and $1+8 = 9$. Try any starting number , you always end up at $9$ (or $0$ if the digits are equal). Why? Because the difference of a number and its reverse is a multiple of $9$.

$1089$ trick. Pick any $3$-digit number whose first and last digits differ by at least $2$. Reverse it. Subtract smaller from larger. Reverse the result and add. You always get $1089$. Try $321$:

$321 - 123 = 198; \quad 198 + 891 = 1089.$

This works for the same digit-sum reason but with two reversals.

Worked examples

Example 1. Find the missing entries in this $3 \times 3$ magic square (magic sum $15$):

$\begin{array}{|c|c|c|}\hline 8 & ? & 6 \\ \hline ? & 5 & ? \\ \hline 4 & ? & 2 \\ \hline\end{array}$

• Top middle: $15 - 8 - 6 = 1$.
• Bottom middle: $15 - 4 - 2 = 9$.
• Middle left: $15 - 8 - 4 = 3$.
• Middle right: $15 - 6 - 2 = 7$.

Example 2. Verify $9 \times 142857 = ?$ does NOT have a simple cyclic pattern.

• $9 \times 142857 = 1{,}285{,}713$.
• $7$ digits , no longer fits the $6$-digit cycle (because $9 > 7$, you have crossed beyond the $7 \times 142857 = 999999$ boundary).

Example 3. Try the $1089$ trick on $592$.

• $592 - 295 = 297$. (Yes, $592$ minus its reverse, $295$.)
• Reverse $297 = 792$.
• $297 + 792 = 1089$. ✓

Example 4. A magic square's magic sum is $34$. If the numbers are $1$-$16$, this is a $4 \times 4$. Find the diagonal sum if one diagonal is $1, 6, 11, 16$.

• $1 + 6 + 11 + 16 = 34$. ✓

Try it yourself

1. Complete this $3 \times 3$ magic square with magic sum $15$: only $2$, $7$, $6$ in top row, $4$, $\_, 8$ in bottom row.
2. Compute $11^5$. Is it a palindrome?
3. Verify $4 \times 142857$ is a cyclic rotation of $142857$.
4. Try the $1089$ trick on three different $3$-digit numbers and check.
5. The magic sum of a $4 \times 4$ magic square using $1$-$16$ is $34$. Show your reasoning.
6. Show that the difference of any $3$-digit number and its reverse is a multiple of $99$.
7. Find a $4$-digit palindrome divisible by $7$.
8. List all $3$-digit palindromes divisible by $11$. (Hint: every $3$-digit palindrome has form $aba$.)

Activity

Make your own magic. Try to build a $4 \times 4$ magic square using the numbers $1$ through $16$. There are several known constructions , one famous one is Albrecht Durer's square from the engraving Melencolia I ($1514$). Look it up online; the bottom row even contains the year $1514$. Once built, verify that all rows, columns, both diagonals, and even the four corners sum to $34$.