Okay, now I’ve figured out how to do it. Now I’ve just got to figure out why it works.
Turns out for some reason when you subtract a scrambled version of a four digit number from the four digit number, at least assuming the answer is positive, the digits add up to a multiple of nine. So given the three digit number, you just figure out what needs to be added in in order to make the four add up to a multiple of nine.
-FrL-
[spoiler]Oh I see. Say the original number is abcd. You scramble it into efgh. Now, you take abcd - efgh. This is the same as 1000a + 100b + 10c + d - 1000e - 100f - 10g - h. Now, either e=a, f=a, g=a, or h=a. Whichever the case may be, once you have subtracted the appropriate e, f, g, or h term from the a term in the above expansion, you end up with a value equal to 9na for some n. For example, 1000a - 1000a = 9 * 0 * a. 1000a - 100a = 9 * 100 * a. And so on.
What is true for a is true for each of a, b, c and d. From this, then, we see that the result of the subtraction must be a multiple of nine (since it is expressible as the sum of four multiples of nine.)
The digits of any multiple of nine will add up to some multiple of nine. So any one of those digits can be reconstructed from the other three, by adding the three, and taking the difference between that sum and the next multiple of nine.
You can’t circle a zero because if the other three digits add up to a multiple of nine themselves, this is compatible with the remaining digit being either a zero or a nine. By disallowing the circling of a zero, you guarantee that if the three add up to a multiple of nine, then the remaining digit was a nine and not a zero.[/spoiler]
-FrL-
Write your own long-winded story to go with this, but the answer that needs to come up on the calculator window is 71077345. Turn the calculator upside down, and it reads “ShELLOIL”.
===
Not really a trick - or could be worked into one, somehow - but a cute coincidence anyhow: ask anyone for their favorite number between 1 and 9. Multiply that by 9, and multiply that by 12,345,679. (Note: the digit ‘8’ is missing. I don’t know how this works, but it does.)
For example, 27 x 12,345,679 = 333,333,333. It just works, is all.
12345679 * 9 = 111,111,111 so you’re saying pick a number times 111,111,111.
Here’s a good trick to teach babies basic math (my favorite to perform):
Find a baby with 3 pieces of candy and say “OK, you have 3 pieces of candy. If I take away two, how many will you have left?” If they are smart, they will hold up one finger. Then you say, “That’s right!” and do it. Now you have two pieces of candy. Score!
Another one is to find out what Bigfoot wears, and the answer works out to 53045618. 
Pick any three digit number. (for example 578)
Now copy it, to make a six-digit number. (i.e. 578578)
Divide by 7.
Now divide what you get by 11.
Now divide by 13.
Came in here to suggest the same thing. I think it does no harm to introduce from an early age that mathematics isn’t just about numbers. Ask your class to colour ‘both’ sides of a mobius strip they make for themselves and see if they notice anything odd might be amusing also.
I don’t expect they’ll learn anything much about non-orientable surfaces but they might get a glimpse that maths is a big and creative topic and that a sense of ‘play’ is important to the process.
I’ve done the whole mobius strip tricks thing with other adults (normally those who wonder why the hell I did maths in university) as a counter-example to all the soulless number crunching they associate with maths. It generally works well at getting that point across.
Take any page of a monthly calendar, and ask someone to draw a (square) box around any 9 dates, without showing you which ones they’ve chosen. (It’ll have to be a 3 x 3 square.) Ask them to tell you just the first of the dates. You immediately announce the sum of all the dates in the square (which someone can check with a calculator).
The way you do this is to take that first date, add 8, then multiply by 9. (Or, equivalently, first multiply by 9, then add 72, since 9(n+8) = 9n + 72.)
Prepare ahead of time a square array of numbers (in my example, I’ll use a 5 x 5 array, but you can make it any size as long as it has the same number of rows as columns), by writing any numbers across the top, and any numbers down the side, and then filling in the array by making each entry = the sum of the number at the top of its column and the number at the beginning of its row. Fow example,
1 2 3 4 5
0 1 2 3 4 5
2 3 4 5 6 7
4 5 6 7 8 9
6 7 8 9 10 11
8 9 10 11 12 13
Now prepare a sheet of paper with just the square array on it (not the numbers across the top & sides). Also, ahead of time add up all the numbers across the top and down the side, write this total (in my example, 1+2+3+4+5+0+2+4+6+8=35) on a slip of paper or index card, and put it in a sealed envelope, and leave it lying on your desk or give it to a student and tell them not to open it until you tell them to.
Now, hand the array of numbers to someone, and tell them to pick any number and circle it. Then, cross out all the other numbers in that same row and all the other numbers in that same column.
Now hand it to someone else, and ask them to pick any of the remaining numbers (that haven’t been circled or crossed out), circle it, and then cross out every other number in its row and column.
Keep doing this until all the numbers are either circled or crossed out. (Each row and each column will have exactly one circled number in it.)
Now, ask someone to add up all the circled numbers and tell you the total.
Ask someone else to open the sealed envelope and tell you what number is inside. They should be the same.
(This trick, and the previous one, I got from books by Martin Gardner.)
Square one.
This one is kind of hard to explain. Have the kids hold up their hands in front of them and give each finger a number so one, two three, four, five is pinky, ring finger, middle, index and thumb and so on. Now you can do the basic 9 times. 9 x 2? Put down the second finger, you get 18. 9 x 5, the left thumb goes down, 45. 9 x 9, 81.
Here’s an interesting one dealing with 7.
1/7 = 0.142857…
2/7 = 0.285714…
3/7 = 0.428571…
4/7 = 0.571428…
5/7 = 0.714285…
6/7 = 0.857142…
Fun Websites
You just blew my mind.
I have to discover what general principle this is an instance of.
-FrL-
Well, if you really want the explanation…
Try dividing 1/7, 2/7, 3/7, etc. by hand. In each case you get a repeating decimal with a period of 6 (i.e. that’s how long the string of digits that repeats is). With a denominator of 7, 6 is the largest possible period, because when you’re dividing by 7, there are only 6 possible remainders. After you’ve gotten all 6 of them, you have to get one you’ve seen before, and that’s where the repeating begins. When dividing 1/7 vs. 2/7 vs. 3/7 etc., you get the same sequence of remainders, but starting at a different point in the sequence.
Other decimals that have the largest possible period are those with denominators of 17 (16 digits repeat), 19 (18 digits repeat), 23, 29, 47, 59, 61, and 97. And so a similar phenomenon happens when you look at fractions with these denominators: the decimal expansions are the same, only starting at different places in the repeating sequence.
For example, 1/17 = 0.0588235294117647repeating,
2/17 = 0.1176470588235294repeating,
3/17 = 0.1764705882352941repeating,
4/17 = 0.2352941176470588repeating,
etc.
16/17 = 0.9411764705882352repeating
(Source: Chapter 2 of Excursions in Mathematics, by C. Stanley Ogilvy)
Thanks, that frames the issue well for me. I’ll do some henscratching to figure out why it is that “you get the same sequence of remainders” for each of the divisions in question. (Which is, in any case, the thing I found mysterious in the first place.)
-FrL-
After a minute’s thought, I think I get it:
In any division, given the divisor, one remainder determines what the next remainder will be.
For some divisions, the period is as large as it can be, and is produced by remainders equal to every number less than the divisor.
This means each remainder must occur at least once in the sequence producing the period.
No remainder can occur more than once in that series, since this would mean the successor remainder in each case would also be the same, and so on, meaning the period is actually smaller than is presently hypothesized.
So each remainder occurs exactly once.
Since a given remainder is always followed by the same successor remainder, it follows that the remainders come in a fixed series, and that this series must loop around since the period must repeat.
Well anyway, that wasn’t at all rigorous, but I think I’ve got the idea. Thanks!
-Kris
There’s this old trick. Take any 3-digit number. Reverse it. Subtract the smaller from the larger. Reverse the result (if it’s only 2 digits, add 0 to the front). Add the result and its reverse. The final answer is always 1089. (It works in other number bases, too. Base 8 gets you 1067, base 16 gets you 10EF.)
To add to the presentation, prepare beforehand by writing 1089 on your arm with bar soap so the digits read vertically down, starting with your wrist. Have the kid write the final answer on a piece of paper and fold it up. You burn the piece of paper to ashes. Rub the ashes on your arm, and they’ll adhere to the soap.
Alternatively, write 6801 on your arm with the soap. Look puzzled for a second when it shows up, then position your arm hand down.
I read the soap thing in an old book of tricks. The reversal is my idea.
Here’s the reason that number happens to work that way.
1/9 = .1111111…
1/9*1/9 = 1/81 = .012345679012345679…
Curses! I wrote a long post thanking folks specifically for their tricks. It’s hamsterchow. So, thanks in general! A quick note: in the second trick, the periods are just there for spacing. If you want to figure out how the trick is done, here’s a hint: if a kid chose the numbers I gave in the example, I would fill the blank with 673, and then I could immediately write down the answer to the problem (784 x 326) + (784 x 673).
Daniel
Most people are familiar with the rule for divisibility by 9. Less well known is the rule for divisibility by 11: add and subtract alternating digits, see if the result is divisible by 11.
so 12,345,670, we get 1-2+3-4+5-6+7-0 = 4, which is not divisible by 11. But add a 7 on the end, and we get 123,456,707, which is 11 * 11,223,337.
Tell 'em to:
[ol]
[li] Pick a number between 1 and 10 (but don’t say it aloud)[/li][li] Double it[/li][li] Add 8*[/li][li] Divide by two[/li][li] Subtract the number originally chosen[/li][/ol]
The answer will always be half of what you tell them to add in step 3–which makes it fun, because you can repeat the trick using different numbers.
*Any even number should work, but as I’m sure you suspect, keeping it small (within 1-10) usually works best for mental calculations at that age.
[QUOTE=Civil Guy]
Write your own long-winded story to go with this, but the answer that needs to come up on the calculator window is 71077345. Turn the calculator upside down, and it reads “ShELLOIL”.
Yep I used that one as a kid in the 70s. 142 Arabs fight 154 Jews over 69 oil wells for 5 days. Who wins? (14215469*5 = ‘ShELLOIL’)
We used to check a (8 digit) calculator for ‘accuracy’ by entering
1.1111111 * = (*= is the shortcut on most calculators to square the number).
the answer is 1.2345678