## A Couple of Mensa Puzzles

Mensa's Number Puzzles for Math Geniuses by Harold Gale has this puzzle:

Number Puzzle 58

Place six three digit numbers of 100 plus at the end of 685 so that six numbers of six digits are produced [like 685123]. When each number is divided by 111 six whole numbers can be found.

Of course there are eight such numbers. The first is greater than or equal to 685100. Dividing that by 111 gives us 6172.072. . . So the first number of our answer should be 111x6173 or 685203. The next answer is 111 more than that (111x6174=685314). Then 111 more than that, etc. The answers:

685203, 685314, 685425, 685536, 685647, 685758, 685869, 685980.

The answers in the back of the book leave out the first and last of these answers.

This puzzle is similar to ones like this:

How many three digit numbers are divisible by 17?

Well, the first one is greater than 100. 100/17=5.88235. . . So the first is 6x17=102. We don't need to list these three digit numbers. The last one is less than 1000. 1000/17=58.8235. . . So our three digit numbers are 6x17, 7x17, 8x17, . . ., 58x17. There are 53 such numbers.

See A Simple Number Puzzle, which is a puzzle for kids (grade school).

The same book has this puzzle (#166): "The contents of each box has a value. The total of the values is shown alongside a row or beneath a column. Which number should replace the question mark?" This is simple algebra: A+B+2C=146 and 2A+B+C=147. We want to get rid of the C's. So, 4A+2B+2C-A-B-2C=294-146, and 3A+B=148, our answer, not a very difficult puzzle. The book gives 146, a typo.

The book further says that A=38, B=34, and C=37. Of course, it is impossible to deduce this from the puzzle, as we have two equations and three unknowns (we have five equations, but most are duplications). Anyway, using these three values, we get 148, not 146.

The first such puzzle (#6) in the book is this (on the right). Similar algebra is not necessary here. Just notice that 48+46+70+54=54+?+54+48, giving us our answer: 62. That method may be simpler than the author intended. Of course, "math geniuses" should be on the lookout for such shortcuts.

I received email point out that the above puzzle (puzzle #166) is even easier than I thought. The row and column which show no sum contain the same symbols, and so can be subtracted out when we use the simpler method above (which I showed for puzzle #6). There are math geniuses everywhere. The same email claimed that puzzle #166's solution of A=38, B=34, C=37 is actually the only solution. But even if we restrict ourselves to positive integer values for each symbol, there are still 48 different solutions (C is any integer from 1 to 48).