3/26/2023 0 Comments Simple math puzzle![]() ![]() It does so in some cases when there are more than four digits. As we shall see below, it could have ended in a repeating cycle where it bounced between two or three different numbers or even more. But even if 6174 were the only nontrivial solution to such subtractions, that does not mean that our procedure has to end in it. Not only that, we would also have to show that other four-digit numbers with two or three repeated digits do not have any solutions. So we would have to try 23 other cryptarithms such as SETA − ATES = TEAS or SEAT, etc., to show that they have no solution. For starters, there are 24 ways that four different digits can be ordered, all of which would yield the same Kaprekar subtraction. We have shown that at least one four-digit number maps to itself using the Kaprekar procedure, but there could conceivably have been others. This only partially answers our previous question about why this occurs. You can actually solve the cryptarithm using conventional methods ( Douglas Felix detailed one way of solving it) and confirm that 6174 is the only solution. Knowing that the only other four-digit string that yields itself is 0000, which is ruled out in this case, we expect there to be a single solution, 6174. The order of the digits matches that of the number 6174, so we expect 6174 to be a solution. The number EAST will generate the subtraction SETA − ATES to yield EAST again. ![]() Observe that the above subtraction problem exactly reflects the procedure that we have been carrying out above. How many solutions do you expect it to have? Here is a cryptarithm (digit substitution) puzzle that may help (incidentally, all these words are acceptable in Scrabble). The maximum number of steps to reach 6174 is seven, which is required by several numbers (1040 is an example).ĭ. ![]() The other 9,990 numbers all result in 6174. As for exceptions, the only four-digit strings that do not end in 6174 with the Kaprekar procedure are the ones with identical digits: 0000, 1111, 2222, etc., for which the procedure ends with 0. Let’s save the first question for the end of this section. ![]() Explore further: Why does this occur? Can you find exceptions? Can you find numbers that require more steps than any of our examples? Keep generating numbers until something interesting happens: 6174 repeats forever.ī. Follow the same procedure to generate sequences starting with the following numbers. The next number in the series is obtained by subtracting the second number in parenthesis from the first. Find the next three numbers in the sequence: 5634 (6543, 3456), 3087 (8730, 0378), 8352 …įor each number in the series, the first number in the parentheses is obtained by arranging the digits in descending order, and the second by arranging them in ascending order. Let’s first learn more about Kaprekar’s constant in this four-minute video:Ī. Since you’ve already done the work of trying to discover these constants, let’s sit back and experience their magic. The first is an interesting curiosity in recreational number theory, while the second is a universal constant central to many chaotic phenomena in the real world. The second, ∂ = 4.6692016…, was discovered in 1975 by the mathematical physicist Mitchell Feigenbaum with the aid of an HP-65 programmable calculator.īoth of these constants remain somewhat mysterious. Kaprekar through pen-and-paper arithmetic explorations. The first constant, 6174, was discovered in 1946 by the Indian mathematician D.R. Readers found that the procedures ended in a repeating cycle - either in a single number (a “cycle” of one) or in a cycle of two or more numbers. In the April Insights puzzle, I tried to guide readers down a path that might be best described as “experimental mathematics.” The goal was to rediscover two constants by iterating simple arithmetic procedures. ![]()
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