Trick 1Start with a number and count the number E of even and the number O of odd digits. Write them down next to each other following by their sum E + O. Treat the result as a new number and continue the process. In this example, iterations converge very rapidly. Moreover, in just a few steps they reach the number 123 which has exactly 3 digits of which 1 is even and two are odd. Therefore, applying the computations to 123 produces the number 123 itself, such that further iterations become really mindless. Of course we can always start with another number. (In the applet below, most of the numbers are clickable so that you can change their values. The starting number may be looked at as a string of individual digits or as a long number depending on whether the Autonomous digits button is checked or unchecked.) Trick 2 Every 4th year is leap unless its divisible by 100. This is only true with one additional caveat: among those divisible by 100, the ones that are divisible by 400 are still leap. As every one can see, this adds to the argument that calendars are deeply involved with mathematics. Or, perhaps, its the other way round. Below there is a Java monthly calendar that may be used to discover some simple mathematics built into calendar tables. Due to a bug in (my) Java libraries, the calendar only works for the years between 1970 and 2037 (inclusive). Its more than enough to make a few math discoveries. This is how you use the applet. When the second button on the left reads Set, drag the cursor inside the table and select a square array of dates. Once you are satisfied with your selection, click the button. The label will change to Reset. Now pick up dates inside the select area by clicking on them. You should (and in fact only allowed to) select one date in each row and in each column. Sum up selected dates. The sum that also appears in the lower right corner does not depend on selection of dates inside the square but only on the square itself. You can check this fact by pressing the same button, which now reads Reset. See if you can verify or even prove this. Trick 3 Start with a number and count the number E of even and the number O of odd digits. Write them down next to each other following by their sum E + O. Treat the result as a new number and continue the process. In this example, iterations converge very rapidly. Moreover, in just a few steps they reach the number 123 which has exactly 3 digits of which 1 is even and two are odd. Therefore, applying the computations to 123 produces the number 123 itself, such that further iterations become really mindless. Of course we can always start with another number.(In the applet below, most of the numbers are clickable so that you can change their values. The starting number may be looked at as a string of individual digits or as a long number depending on whether the Autonomous digits button is checked or unchecked.) Trick 4 Math TelepathyAfter reshuffling a particular digit d may or may not change its position. In the decimal representation of the chosen number, d was a coefficient of a power of 10, say 10k. After reshuffling, the power may be different: 10m. Now, if we take the difference of two numbers term by term, i.e. seeking to subtract two terms related to the same digit even if its position changed, well get the sum of differences, like d(10k - 10m). Note the any power of 10 equals 1 modulo 9. This implies that every term d(10k - 10m) equals 0 modulo 9. Therefore, the whole difference of the two numbers (the chosen and the reshuffled) equals 0 modulo 9. When you communicate the computer all the digits, except the selected one, the computer finds the sum modulo 9 of the presented digits and subtracts the result from 9. This is exactly the digit you have omitted.(Instead of subtracting the smaller number, one can subtract their digits in turn, or their sum just once. The result is a multiple of 9 and the trick works as before. In case of 2-digit numbers the triack has a different computer incarnation.) Trick 5 Magic in a SquareSimple as the trick is, it was described by [Rouse Ball, p. 325] and referred to by [M. Gardner, p. 3]. Martin Gardner writes: The success of the trick depends, of course, on the spectators inability to follow the procedure well enough to guess the operating principle. He then selfishly adds: Unfortunately, few spectators are that dense.Every location in the square is determined by its row and column numbers. The column is indicated directly on the first response, after which the columns of the square are swapped with the corresponding rows. Answer to the second query therefore determines the row. To confound the mildly dense spectators, computer, before displaying the numbers, randomly reshuffles them in every column. Trick 6 No Questions AskedAsk A to take any number of counters that he pleases: suppose that he takes n > 1 counters.Ask someone else, say B, to take p times as many, where p is any number you like to choose.Request A to give q of his counters to B, where q is any number you like to select.Next, ask B to transfer to A a number of counters equal to p times as many counters as A has in his possession. Then there will remain in Bs hands q(p + 1) counters; this number is known to you; and the trick can be finished either by mentioning it or in any other way you like.The reason is as follows. The result of operation (ii) is that B has pn + q counters, and A has n - q counters. The result of (iii) is that B transfers p(n - q) counters to A; hence he has left in his possession (pn + q) - p(n - q) counters, that is, he has q(p + 1) .For example, if originally A took any number of counters, then (if you chose p equal to 2), first you would ask B to take twice as many counters as A had done; next (if you chose q equal to 3) you would ask A to give 3 counters to B; and then you would ask B to give to A a number of counters equal to twice the number then in As possession; after this was done you would know that B had 3(2 + 1), that is, 9 left.This trick (as also some of the following problems) may be performed equally well with one person, in which case A may stand for his right hand and B for his left hand. Trick 7 ExplanationAssume at a certain stage of the process we have a sequence S of integers, S = {A1, A2, ..., An}. Let P stand for the product of all integers present, each increased by 1 less 1: P = (A1+1)(A2+1)...(An+1) - 1. When two numbers A and B are replaced with (A•B + A + B) the product P does not change. Indeed, it loses two factors (A+1)(B+1) but gains another one (A•B + A + B + 1). The two are of course equal.It then follows that P is an invariant quantity that does not change in the course of the game. It could be computed at the very beginning. For example, if we are given the sequence 1, 2, 3, 4, 5, 6 then P = 7! - 1 which you can easily verify. Trick 8Gergonnes Magic Trick The numbers from 1 through 27 are displayed below in three rows of nine numbers each. Select one of those numbers and reply truthfully to three computer queries Trick 9 1089 and a Property of 3-digit NumbersNumber 1089 is a centerpiece of a curious mathematical trick used to stun the uninitiated with the performers math prowess. I was reminded about it watching David Achesons video. David is the author of highly recommended book 1089 and all that.Take any 3-digit number, say, 732 and write it backwards: 237. Subtract the smaller of the two numbers (237 in our case) from the larger (732). With our selection, we obtain the number 495. Write this one backwards too and compute the sum: 495 + 594. Here comes the surprise: regardless of your original selection, the final result will always be 1089!I am going to explain shortly why this is so, but first either give it a thought or experiment with the applet below. The blue numbers are clickable. The digits increase if clicked to the right of their vertical axis, and decrease if clicked to the left of the axis. Trick 10 Bachets Magic TrickLet N be an integer. This is the lone number at the bottom of the applet below. It can be modified by clicking on it. Currently it is 4. For practical reasons, it can change between 3 and 6, inclusive.The integers from 1 through N•(N+1) are split into pairs. Select one pair (not just two integers but a pair among the pairs of integers the applet displays). Remember this pair, think hard of it with complete concentration, click the Proceed button. The same numbers will now be arranged in a N×(N+1) array. By just pointing to the two rows in which your selected numbers are located (if the numbers are in the same row, select it twice) you give your computer enough information to guess your numbers. Try it!
Posted on: Wed, 06 Aug 2014 03:06:45 +0000
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