Back in January of 2012, I wrote about the Chinese Remainder Theorem. Also, Martin Gardner taught the basics well in his book Aha!: Insight, including a trick where you can determine someone's secretly chosen number between 1 and 100 just from hearing the remainders when divided by 3, 5 and 7. Going over that post again, I've developed a few improvements to this trick that make it seem much more impressive, and maybe even easier to do.
The first problem is getting the remainders. With a standard calculator, it's not easy to do. The answer here is to simply have them divide the number by 3, 5 and 7, and have them tell you ONLY the number after the decimal point. Using the amount after the decimal point, you can work out the remainder. When dividing by 3, there are 3 possibilities for numbers after the decimal point:
- Nothing after the decimal point: Remainder = 0
- Number ends in .3333...: Remainder = 1
- Number ends in .6666...: Remainder = 2
- Nothing after the decimal point: Remainder = 0
- Number ends in .2: Remainder = 1
- Number ends in .4: Remainder = 2
- Number ends in .6: Remainder = 3
- Number ends in .8: Remainder = 4
- Nothing after the decimal point: Remainder = 0
- Number ends in .14: Remainder = 1
- Number ends in .28: Remainder = 2
- Number ends in .42: Remainder = 3
- Number ends in .57: Remainder = 4
- Number ends in .71: Remainder = 5
- Number ends in .85: Remainder = 6
The other improvement involves the process itself. Have them start by dividing their number by 7 and telling you the numbers after the decimal point. Using the steps above, you can quickly determine the remainder from 0 to 6. If the remainder is between 0 and 5, you can remember them by touching that many fingers of your left hand to your pant leg (or table, if present). For a remainder of 6, just touch 1 finger from your right hand to your pant leg or table.
What ever the remainder, imagine a sequence starting with this number, and adding 7 until you get to a number no larger than 34. For example, if the person told you their number ended in .42, you know the remainder is 3, and the sequence you'd think of is 3, 10, 17, 24 and 31 (we can't add anymore without exceeding 34). If the remainder was determined to be 4, instead, the sequence you'd think of would be 4, 11, 18, 25 and 32.
Next, ask the person to divide by 5, and tell you the part of the answer after the decimal point. Once you get this number, recall your earlier sequence (which can be recalled via the number of fingers resting on your pant leg or table), and subtract the new remainder from each number, until you find a multiple of 5. For example, let's say that when dividing by 7, their remainder was 4, so the sequence was 4, 11, 18, 25 and 32. Let's say that when dividing by 5, their remainder was 3. Which number from your initial sequence, when it has 3 subtracted from it, is a multiple of 5? is 4-3 a multiple of 5? No. Is 11-3 a multiple of 5? No. Is 18-3 a multiple of 5? YES! Now you have the number 18.
Whatever number you have at this point, is the smallest of 3 possible numbers. The other 2 possibilities are 35 more than that number and 70 more than that number. At this point, you can know that the chosen number is either 18, 53 (35+18) or 88 (70+18). The remainder when dividing by 3 will determine which one of these is the correct answer. For example, if they say their number, when divided by 3, ends in .3333..., you know the remainder is 1. So, run through all 3 numbers quickly and ask yourself which one is 1 greater than a multiple of 3. Is 18 - 1 a multiple of 3? No. Is 53 - 1 a multiple of 3? No. Is 88 - 1 a multiple of 3? YES! Therefore, their number must be 88.
There are certainly other approaches. In fact, a magician's magazine called Pallbearer's Review once presented this trick as a challenge, asking their readers to supply their methods. They received a wide variety of entries and many approaches. Most of them involved far more difficult methods than the above approach, which I prefer for actual performance.
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