TRICKS FOR ARITHMATIC
Added on: 13th Dec 2015
THE MAGIC STRING
Imagine that you tied a string around the equator of the Earth so tight
that you couldn’t even fit a razor blade underneath. Now let’s
imagine that we lengthen the string by only 1 metre. Of course we would
now have some slack around the equator, but how much? It’s hard to
believe but the answer is that the string would now clear the Earth by
16cm all the way around! If you want a party trick just google the proof.
It will fit onto a napkin.
THE COIN SORTER
Lay out a bunch of coins on the table and tell your friend to blindfold you.
Ask him how many of the coins are facing heads up. Whatever number
he tells you, flip that many coins over (any coins) and move them to a
separate pile. You will now have two piles with the same number of
heads and tails and your friend will think you are a wizard after he counts
them! To add some drama, pretend to select the coins you flip carefully.
Why does this work? It’s maths!
FIGURE OUT THE LAST NUMBER OF ANY BARCODE
The last digit in any barcode (the one that is apart from the rest and
not under the bars) is actually used by the computer to check and
make sure it reads the numbers right. Impress your friends by being
able to “guess” these! Starting from the right add every odd digit three
times and every even digit once. Then subtract the last digit of the
total from 10. Here is an example:
For 03600029145 you should calculate something like this:
5+4+1+9+2+0+0+0+6+3+0+
5+1+2+0+6+0+
5+1+2+0+6+0 = 58
10 – 8 = 2
The extra digit would be 2!
CHECK ANY MULTIPLICATION PROBLEM
This makes use of a trick called digital roots. For 2878 x 4902 = 14107956
just do the following:
Find the digital roots of the first number:
2+8+7+8 = 25
2+5 = 7
Do the same for the second and third numbers.
We’ll spare you the time and tell you that they are both 6.
So, take 7×6 (the digital roots of the two numbers you are multiplying)
which equals 42. 4+2 = 6. Since 6 = 6 the math is right!
THE CALENDAR TRICK
Tell your friend to select a square of 9 numbers on any calendar.
For example:
14 15 16
21 22 23
28 29 30
No matter what square he chooses you can quickly tell him what
they all add up to. Just multiply the middle number by 9! 22 x 9 = 198
THE CALENDAR TRICK ON STEROIDS
This time tell your friend to select a 5×4 box around any 20 numbers
on the calendar. All you have to do to figure out what they all add up to
is take the lowest number and highest number and add them together.
Then multiply the answer by 10.
CALENDAR TRICK EXTENDED
The previous two tricks will actually work on any grid of numbers
as long as it is continuous!
THE MONTY HALL PROBLEM
First gaining public attention when it was sent to Ask Marylin
(Marlylin vos Savant’s column in Parade Magazine), the answer to this
statistical anomaly at first caused quite an uproar. Some Phd’s and
mathematicians (even from MIT!) wrote to the magazine in disbelief.
After several months though, with some scientists even designing
computer simulations to prove it, the answer to the Monty Hall Problem
showed itself to be correct. And here is the problem as it was written to
Marylin in 1990: Suppose you’re on a game show, and you’re given the
choice of three doors: Behind one door is a car; behind the others, goats.
You pick a door, say No. 1, and the host, who knows what’s behind the
doors, opens another door, say No. 3, which has a goat.
He then says to you, “Do you want to pick door No. 2?”
Is it to your advantage to switch your choice?
The answer is incredibly that yes, your chances increase if you switch doors.
You’ll have to google it to find all the proofs but a quick way to visualize it
is to imagine not 3 doors but 1 million doors. You choose 1 door and then
the game show host opens all but 1 other door. This time the answer
becomes more obvious. You should definitely switch.
Would you really trust yourself to have picked the right door out
of 1 million? Here is another intuitive explanation offered by Matthew Carlton:
An intuitive explanation is that if the contestant picks a goat
(2 of 3 doors) the contestant will win the car by switching as the other
goat can no longer be picked, while if the contestant picks the car
(1 of 3 doors) the contestant will not win the car by switching.
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