Math you can perform
Tonight's idea
A lot of stage magic is mathematics wearing a cape. It always comes out right because arithmetic has no choice. Tonight you get four tricks you can perform at dinner, plus the explanations the magician keeps to themselves.
Trick 1: 1089
Method. Ask someone to write down any three-digit number whose first and last digits are different, say 732. Reverse it (237) and subtract the smaller from the larger: \(732 - 237 = 495\). Reverse that (594) and add: \(495 + 594 = 1089\). Before they start, you have written 1089 on a slip of paper and sealed it in an envelope.
Why. Call the digits \(a, b, c\) (for 732, \(a = 7\), \(b = 3\), \(c = 2\)), so the number is \(100a + 10b + c\) and its reverse is \(100c + 10b + a\). Subtracting, the \(b\)s cancel and you get \(99(a - c)\). Take \(a\) as the larger. When \(a - c\) is between 2 and 9, the result is one of 198, 297, 396, 495, 594, 693, 792, 891. Look at those: the middle digit is always 9, and the outer digits always add to 9. Reverse any such number and add it to itself and you get \(9 \times 100 + 18 \times 10 + 9 = 1089\). One caution: when the outer digits are 1 apart the difference is 99. Write it as 099 and reverse to 990.
Trick 2: shoe size and age
Method. Say to someone: take your shoe size (whole sizes only, because the size has to fit in the hundreds column), multiply by 5, add 50, multiply by 20, then add 1026 if you have already had your birthday this year, or 1025 if you have not. Finally subtract the year you were born. They tell you the answer, and you announce their shoe size and age. If they say 712, you say "size 7, and you are 12".
Why. Shoe size \(s\): \((5s + 50) \times 20 = 100s + 1000\). Adding 1026 gives \(100s + 2026\). Subtracting the birth year \(y\) gives \(100s + (2026 - y)\), and \(2026 - y\) is their age. The shoe size sits in the hundreds place and the age in the last two digits. This version expires on 31 December 2026; next year change 1026 to 1027.
Trick 3: the mind-reading cards
Five cards, each listing some of the numbers from 1 to 31. The spectator thinks of a number and tells you which cards it appears on. You add the first number on each of those cards and announce their number. This is binary in disguise. The full method is in the sheet Counting to 31 on one hand, including the lists for making the cards.
Trick 4: the predicted sum
Method. Ask for any five-digit number, say 47381. Write a prediction, fold it, and hand it over. Now play two rounds: in each round they write another five-digit number under the last, and you write one beneath it. After the two rounds there are five numbers in a column. Ask them to add them up. The total matches your prediction.
The secret: your prediction is their first number with a 2 put in front and 2 subtracted: \(247381 - 2 = 247379\). In each round, you write the number that makes their new number up to 99999, digit by digit (each of your digits is 9 minus theirs). So if they write 52917 you write 47082.
Why. Each round adds exactly \(99999 = 100000 - 1\) to the total. Two rounds add \(200000 - 2\). So the total is their first number plus 200000 minus 2, which is what you wrote down before anything happened.
Watch one: the predicted sum, in full
They write 47381. You secretly compute 247381 − 2 = 247379 and write it as your prediction. Round one: they write 52917; you write 47082 (check: 5+4, 2+7, 9+0, 1+8, 7+2 all make 9). Round two: they write 30846; you write 69153.
| Who | Number | Pair |
|---|---|---|
| They | 47381 | |
| They | 52917 | 52917 + 47082 = 99999 |
| You | 47082 | |
| They | 30846 | 30846 + 69153 = 99999 |
| You | 69153 | |
| Total | 247379 | 47381 + 99999 + 99999 |
Open the envelope.
Performance note: if the spectator writes a number starting with 9, your reply starts with 0, which looks odd; ask for numbers "with no 9s" at the start, casually, and nobody will notice.
Your turn
- Run the 1089 trick on 851 and on 403 (careful with the 099 case). Then look at the list 198, 297, …, 891. What do the outside digits add to? Why does that make the second step always give 1089?
- Do the shoe size trick on yourself. Then work out what answer a person with shoe size 11 who is 45 and has had their birthday would get. What goes wrong with shoe size 5½?
- Someone starts the predicted-sum trick with 90210. What is your prediction? They then write 18374. What do you write?
- Perform at least two of the tricks on someone at home tonight. Write one sentence about what they said afterwards.
Stretch: invent one
Here is a fact from the clock arithmetic sheet: any number and the sum of its digits leave the same remainder when divided by 9. So a number minus its digit sum is always a multiple of 9. Design a trick around it. One starting point: ask someone to write a big number, add its digits, subtract that from the number, then cross out any one non-zero digit of the result and read you the rest. You name the crossed-out digit. Work out how you do it, write the full method and the reason, and test it twice.
Answers
- 851 − 158 = 693, 693 + 396 = 1089. 403 − 304 = 099, 099 + 990 = 1089. In every number on the list the outside digits add to 9 and the middle digit is 9. Reversing swaps the outside digits, so adding gives 9 hundreds, 18 tens and 9 units: \(900 + 180 + 9 = 1089\).
- Size 11, age 45, birthday passed: \((55 + 50) \times 20 = 2100\), \(+ 1026 = 3126\), minus birth year 1981 gives 1145: size 11, age 45. Size 5½ gives \(100 \times 5.5 = 550\) plus the age, so the shoe size and age blur together (a 12-year-old would get 562, which reads as size 5, age 62). Round half sizes up or down first.
- Prediction: 290210 − 2 = 290208. Reply to 18374: 81625.
- Any honest sentence.
Stretch. The result after subtracting the digit sum is a multiple of 9, so its digits add to a multiple of 9. Add up the digits they read you, and the crossed-out digit is whatever brings that total up to the next multiple of 9. Example: 7482 has digit sum 21, and 7482 − 21 = 7461. They cross out the 6 and read "7, 4, 1". That sums to 12, the next multiple of 9 is 18, so the missing digit is 6. Asking for a non-zero digit avoids the ambiguity between 0 and 9.