Row, Row, Row Your Square 1. The key to solving problem 1 is recognizing that in each of the four rows of numbers that include the center square, the sum of the first and the last numbers must equal 10.
2.
3. Note that in each of the four rows of numbers that include the center square, the first and last numbers must add up to 10, just as in problem 1.
4. You can build a multiplication magic square by subtracting 1 from each entry in the magic square from problem 1, then raising 2 to each of those powers.

          2^{7}   2^{0}   2^{5}     2^{2}   2^{4}   2^{6}     2^{3}   2^{8}   2^{1}     
In the resulting square, the product of every row is 2
^{12} = 4,096. But you can find a solution with a smaller product by using the "123" square from problem 2. First, subtract 1 from every entry in the "123" square to make a "012" square (A, below). Raise 2 to each power in square A to get square B. Then rotate square A as shown in C, and raise 3 to the powers in that square, as shown in square D.
Now multiply the corresponding numbers in square B and square D.
None of the numbers in the resulting square are the same, because each entry is a different combination of powers of 2 (always even) and 3 (always odd). The product of every row, column, and main diagonal in this square is 2
^{3} x 3
^{3} = 216. This result is related to a mathematical topic called Latin squares, which scientists use to systematically test combinations of conditions, such as drug interactions.
Can You Digit? 1. The smallest solution is shown here. You can switch the two digits in the tens place or the two digits in the ones place without fundamentally changing the solution.
2. The largest solution: 43 + 5,978 = 6,021.
3. The smallest solution: 437 + 589 = 1,026. The largest solution: 843 + 759 = 1,602.
4. 0 + 1 + 2 + 4 + 5 + 7 + 8 + 9 = 36.
5. There are 13 solutions for multiplying a onedigit number by a fourdigit number to equal a fivedigit number. The smallest such solution is 4 x 3,907 = 15,628; the largest such solution is 7 x 9,403 = 65,821. Note that the multiplicands of both equations consist of the same digits. There are four other similar solutions: 4 x 7,039 = 28,156; 7 x 3,094 = 21,658; 7 x 4,093 = 28,651; and 7 x 9,304 = 65,128. There are eight solutions for multiplying a twodigit number by a threedigit number to equal a fivedigit number. The smallest such solution is 39 x 402 = 15,678; the solution with the largest product is 63 x 927 = 58,401. There are two similar solutions with the product 16,038: 27 x 594 and 54 x 297. I found the solutions by writing a spreadsheet that computes all possible products of two numbers within the desired ranges and flags equations that have no repeated digits.
Factor Fiction 1. 1,000 can be factored into 8 and 125. 10,000 can be factored into 16 and 625.
2. In general the only way to factor a 10
^{n} into two numbers, neither of which contains a zero, is to split it into 2
^{n} and 5
^{n}, since any number that has both 2 and 5 as factors will end in a zero. The smallest power of 2 or 5 that contains a zero is 5
^{8} (390,625), so the smallest factor of 10 that cannot be factored into two zeroless factors is 100,000,000 (2
^{8} x 5
^{8}).
3. Thanks to puzzler Harry Nelson for contributing this problem. He verified that the only values of
n up to 512 for which neither 2
^{n} nor 5
^{n} contains a zero are 0, 1, 2, 3, 4, 5, 6, 7, 9, 18, and 33. The best answer is 10
^{33} (8,589,934,592 x 116,415,321,826,934,814,453,125). It seems extremely unlikely that there are any larger answers, because the probability that a given number contains a zero increases as numbers get larger, but no one has proved it yet.
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