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Math Help - Math Tricks of the Trade

  1. #1
    A riddle wrapped in an enigma
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    Math Tricks of the Trade

    Do you have any interesting (amazing) math tricks that can be demonstrated without too much setup...something simple to use in a classroom setting.

    Here's one I ran across a few years ago. At no time do you see what the student is writing.

    1. Have a student choose a 4, 5, or 6 digit number. (Actually, any size will do, but it gets more complicated with lots of digits)

    2. Have the student rearrange these digits to form a different number.

    3. Have the student subtract the smaller number from the larger number.

    4. Have the student circle a non-zero digit in the remainder. (It must be a non-zero digit and you may have to define the word "remainder")

    5. Have the student tell you the digits he DID NOT circle.

    6. You can then tell him the digit he DID circle.

    Here's how it works:
    1. Suppost the student arbitrarily chooses: 25781

    2. Student arbitrarily rearranges to: 75128

    3. Student finds positive difference to be: 49347

    4. Student arbitrarily circles the digit: 3

    5. Student tell you the digits he did not circle, namely: 4947

    6. You immediately tell him that he circled the digit: 3

    How did I know it was 3 ?
    Last edited by masters; August 2nd 2008 at 11:48 AM.
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  2. #2
    A riddle wrapped in an enigma
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    Explanation: Just highlight between the <> to see how it's done.

    <The sum of the digits in the remainder turns out to always be a multiple of 9. So, when you are given the non-circled digits, just add them mentally and see what you have to add to that total to make it a multiple of 9,,In the example above, 4+9+4+7 adds up to 24. You must add 3 to get the next multiple of 9 which is 27. So that's why it's 3.>
    Last edited by masters; August 3rd 2008 at 11:31 AM.
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  3. #3
    A riddle wrapped in an enigma
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    Ok, I can see everybody's gettin' into this, so here's another one.

    Instant sum

    1. Use blackboard or paper and ask the student to pick 4 arbitrary 4-digit numbers.

    2. You write them down.

    3. Now you say that you are going to pick four 4-digit numbers of your own.

    4. You write them down under the first set of four. You now have eight 4-digit numbers.

    5. You ask the student to supply you with one more number.

    6. As soon as you write it down, you instantly produce the sum of all nine numbers. Have the student verify the sum using a calculator.

    Here's how it's done

    1. The first four 4-digit numbers are truely arbitrary.

    2. However, your next four 4-digit numbers must make pairs with the student's numbers so that the sum of each pair is 9999. That's pretty easy to do. Say, the first number was 2438. Then your number would be 7561. You can be sneaky about where you place these numbers to throw off any possibility of someone noticing this pattern.

    3. Once you have 4 pairs of 9999, use the student's last number (the ninth number) as the key.

    4. Here's how the key works. Let's say his number was 4865. You'll want to mentally subtract 4 (which represents the number of pairs of 9999)from the key, and then place a 4 at the beginning of the key. The resulting sum in this case would be 44,861.

    There are variations on this theme. You can have the student pick any number of 4-digit numbers. Doesn't have to be 4. But, if he picks 5, you have to pair those 5 with another 5 in the same way as before to get 9999 for each pair sum. The final answer using 5 pairs and the key (total of 11 numbers) will be found by subtracting 5 from the key and placing a 5 in front of the key.

    You can also do this with any size number 5-digit, 6-digit, n-digit, and the steps are the same.

    This is how it goes using 4-digit numbers

    Student picks:

    4365
    5478
    1357
    3974

    You pick:

    5634
    4521
    8642
    6025

    Ask for one more number. Student picks:

    4866 <==== Use this as the key
    ------
    44,862 <==== Instant Sum
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  4. #4
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    . . . . . Fortune Telling


    1. Ask the student to secretly select any three-digit number, abc.

    2. Have him enter his number twice in succession on his calculator,
    . . forming a six-digit number, abcabc.

    3. From across the room, announce that he selected a very lucky number.
    . . The number is divisible by 7.
    . . Have him divide by 7.

    4. Ask if his zodiac sign is Gemini.
    . . Whatever his answer, explain that you're getting an impression of "twins".
    . . (Hold up two fingers.) .The new number is divisible by 11.
    . . Have him divide by 11.

    5. Have him divide by his original three-digit number.

    6. As he does so, say, "Oh-oh, too bad!"

    7. His final answer is 13.



    . . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Secret

    Writing a three-digit number (abc) twice, we have abcabc.

    Note that abcabc \:=\:1001 \times abc \:=\:7 \times 11 \times 13 \times abc

    . . . . . . See?

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  5. #5
    Senior Member euclid2's Avatar
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    little math trick for classroom setting

    Ok This is a good one for the classroom. Write on the board 1011010 and tell the students/audience that you need to somehow get nine-fifty out of those numbers and you are allowed to moved one of the numbers, only once, but you can move them anywhere (turn them sideways and upside down if you want) Ok, now once nobody had gotten it which is most likely; take the second 1 and turn it sideways to make the letter TO so it will now be 10t010 (in other words- nine-fifty is the same as ten to ten)
    Last edited by euclid2; August 8th 2008 at 04:16 PM.
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  6. #6
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    Hello, euclid2!

    That's good one . . . a groaner!

    I think you need one more "1" in your diagram.
    Code:
               ___         ___         ___
          |   |   |   |   |   |   |   |   |   |
          |   |   |   |   |   |   |   |   |   |
          |   |___|   |   |___|   |   |___|   |

    It can be presented with matchsticks or toothpicks.

    "Move one matchstick and change it to nine-fifty."
    Code:
               ___   ___   ___         ___
          |   |   |   |   |   |   |   |   |
          |   |   |   |   |   |   |   |   |
          |   |___|   |   |___|   |   |___|

    We probably should spread them to display: . 10 \text{ TO }10


    . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    You think that's a groaner? . . .


    Move two matches and show what matches are made of.

    Code:
                   ___     ____    ___
          |   |   |   |   \    /  |
          |   |   |   |    \  /   |
          |___|   |   |     \/    |___


    Solution:
    Code:
                   ___             ___
          |       |   |   \    /  |
          |       |   |    \  /   |---
          |___    |___|     \/    |___

    I warned you, didn't I? . . .

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  7. #7
    Senior Member euclid2's Avatar
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    lol! Yes, absolutely there should be one more 1 in there, great pictures thats what it is meant to look like!
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  8. #8
    A riddle wrapped in an enigma
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    Applying a Fibonacci series

    Here's one that uses a sort of Fibonacci series to determine the result of a sum rather quickly. The setup and explanation seems quite long, but it is easy to accomplish. I didn't want to spare any detail. Try it.

    1. On a worksheet (blackboard) put a column of numbers from 1 to 10 with a dash after each number. This has a dual purpose. It makes sure we have exactly 10 numbers to add, and that we can pick out the 7th number at a glance.

    2. Have someone choose a number between 5 and 15 (actually, you could have them choose any number at all, any size and the trick will work). Jot that number after 1-

    3. Have another person choose another number between 10 and 20 (Here again, you could have them choose any number they want). Jot that number after 2-

    4. Have either person add these two numbers together to get the 3- number and so on, adding each sum to the previous sum to arrive at the next sum just like the Fibonacci sum. Stop at the 10- number.

    5. Immediately produce the sum and say "Ta da!"

    How it's done

    Each term in a Fibonacci series containing 10 numbers is a fraction of the total. It happens that the 7th number is always precisely 1/11 of the final total. No other number in the series provides a consistent result. So you just multiply the 7th number by 11.

    How do you multiply a number by 11 mentally?

    In this example, easy. But if you use larger numbers to start with, maybe a little tougher.

    If you use the restriction placed upon choosing the initial 2 numbers, then the 7th number will be a 3-digit number. Here' s how you multiply a 3-digit number by 11 mentally. I know most of your probably already know how, but indulge me. If you don't know how, you might look here: Math shortcut trick 11 times tables or study these examples.

    Suppose the 7th number was 227.

    1) First, bring down the 7 as the ones digit
    2) Second, add the 2nd and 3rd digit to get 9 as your tens digit.
    3) Third, add the 1st and 2nd digit to get 4 as your hundreds digit.
    4) Fourth, bring down the 1st digit 2 to finish.
    5) Result: 2,497

    Sometimes you have to carry digits as you would in normal addition. For example. Suppost the 7th number was 168.

    1) First, bring down the 8 as the ones digit
    2) Second, add the 2nd and 3rd digit to get 14, but just place the 4 as your tens digit and carry the 1.
    3) Third, add the 1st and 2nd digit to get 7, and then add the 1 that you are carrying to get 8 as your hundreds digit.
    4) Fourth, bring down the 1st digit 1 to finish.
    5) Result: 1,848

    Here's a complete setup and finish:

    1-
    15 <===== first number chosen
    2- 10 <===== second number chosen
    3- 25 <===== Fibonacci sums continue to 10 numbers
    4- 35
    5- 60
    6- 95
    7- 155 <===== this is the number you multiply by 11 to get final total
    8- 250
    9- 405
    10- 655
    ---------
    1,705 <=====155 X 11

    Bring down 5
    5+5=10, jot down 0 and carry 1
    1+5=6 and carry 1 to get 7
    Bring down 1
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