Problem 1 – Part
a) The 'magic' trick involves choosing a single digit number (by
picking an item and using the number of letters in its name),
multiplying by 5, adding 3, multiplying by 2, and then adding a
second single digit number. Then by subtracting 6, you can find the
two original numbers – the tens digit will be the first number, and
the ones digit will be the second number.
From an algebraic
perspective, the whole ordeal boils down to 2(5x + 3) + y, which
simplifies to 10x + y + 6, where x is the first number and y is the
second number chosen. Hence after subtracting 6, it becomes 10x + y,
so as long as x and y are single-digit numbers, this is the same as a
two digit number with x as the tens digit and y as the ones digit.
To explain this to a 6th or 7th grader, I would
walk them through each step...
* pick x
* becomes 5x
* becomes 5x + 3
* becomes 2(5x + 3)
= 10x + 6 – this step would probably take extra time to show how
the two ends up multiplying to both the original 5x and the plus 3,
to become '10x plus 6'. Might have to work through several numerical
examples to help show the distributive property, while pointing out
that number is always six more than the original number times ten.
* becomes 10x + 6 +
y
* then we need to
subtract six to get 10x + y
* 10x + y gives us
our two original numbers – depending on the student's understanding
of place value, this step might also take some extra explaining to
help them see that the tens digit and the ones digit of a two digit
number can also be represented as 10x + y
Problem 1 – Part
b) So if we want to change the 'magic' trick so that we need to add 8
at the end, rather than subtract 6, then we want our final equation
to boil down to 10x + y – 8. Working backwards, then we need the
intermediate equation to be 2(5x – 4) + y, so in the telling of the
trick, we would need to have the person subtract 4 in the middle,
instead of adding 3.
Problem 2 – My own
math magic problem! Blatantly stolen from...
http://www.pedagonet.com/Maths/missingnumber.htm
Have your subject
write down a number, at least four digits long. Then have them add
up the digits in their number, and subtract that amount from their
original number. Now in this new number, they cross out one of the
digits, any digit. Then they tell you what the number is without the
digit they crossed out (the webpage says to have them slowly read you
the digits, but I think we can manage fine if they tell us the
number, plus then it's less obvious that you're doing something with
the digits). In your head, add up the digits off the number they
give you and figure out what you would have to add to get to the next
multiple of nine. That's the digit that they crossed out!
Example... first
number 13579. Digits add up to 25, so we subtract 13579 – 25 to
get 13554.
Now cross off a
digit in 13554 to get 1554. The digits of 1554 add up to 15, so we
would need 3 to get to 18 (the next multiple of nine). Hence your
friend crossed off a 3. You found the secret number! :-)
But why does it
work? (which is not explained on the website I stole the problem
from)
1) Whenever you add
up the digits of a number and then subtract them from the number,
you'll get a multiple of nine. Why? Because of powers of ten and
place value. Let's say we have a number with digits a, b, c, d, e –
then the number could be seen as 10000a + 1000b + 100c + 10d + e. So
if we subtract off the sum of the digits, we get 10000a + 1000b +
100c + 10d + e – (a + b + c + d + e) or 9999a + 999b + 99c + 9d,
which will always be divisible by 9. (I figured this out with help
from google and http://mathforum.org/library/drmath/view/62561.html)
2) Now that you have
a number that's divisible by nine, the sum of its digits will always
be divisible by nine, because of the magic power of nines! I've
known this particular 'trick' for a while, but never really thought
about why it works. So more googling...
http://mathforum.org/library/drmath/view/67061.html
If we have a number
with digits a, b, c, d, e that is already divisible by nine, then by
place value, again we could see the number as 10000a + 1000b + 100c
+ 10d + e, or we could also think of it as (a + 9999a) + (b + 999b) +
(c + 99c) + (d + 9d) + e, or as (a + b + c + d + e) + (9999a + 999b +
99c + 9d). So the original number (which is divisible by nine) is
the sum of the digits plus a number divisible by nine. If we
rearrange the 'equation' to get, we get that the sum of the digits
equals the original number (which is divisible by nine) plus another
number divisible by nine. We can factor out the nine from those two
numbers to see that the sum of the digits must also be divisible by
nine.
3) So we have a
number that we know is divisible by nine, and hence the sum of its
digits is divisible by nine, then if we don't add in one of the
digits (the secret number), we can figure it out by seeing what we
would have needed to still add in order to get a multiple of nine.
Problem 3 – The
alphanumeric puzzle!
I enjoyed doing
these types of problems as a kid in the book -
http://en.wikipedia.org/wiki/Sideways_Arithmetic_from_Wayside_School
The book had some
problems with numbers written as words, where it would work out to
another number (written as a word), but I can't remember any of them
now, so I used the app to come up with...
ONE + THREE = HIPPO
Ok, let me try to
solve it without using the app...
THREE
+ ONE
HIPPO
So first some
observations...
* E + E = O so O is
even.
* There needs to be
a carry over onto the T, so that T + carry = H (because if there was
no carry, then T would equal H, which we can't have). And there also
needs to be a carry over onto the H in three, so that H + carry = I
for the same reason. And in the addition of two numbers, the largest
(and only) carry we can have is one. So T + 1 = H and H + 1 = I, and
thus T + 2 = I. Or we can think of T, H, and I as consecutive
numbers.
Let's try some
possibilities...
* E can't be 9,
because if E was 9, then 9 + 9 = 18, O = 8 and carry the one, then 1
+ E + N = P,
or 1 + 9 + N = P,
but that's 10 + N = P, which would make N & P be the same digit
(and carry the one), but each letter is a unique digit, so E can't be
9.
* What if E equals
8? Then O = 6.
* What if N equals
9, then P would be 8 – but E is already 8, so NOT CORRECT
* What if N equals
7, then P would be 6 – but O is already 6, so NOT CORRECT
* What if N equals
5, then P would be 4, and then 1 + R + 6 = 4 (or rather 14), which
makes
R equal 7. But then
H, I, and T end up being small numbers, which won't result in two
carries.
Hmm... maybe I need
to try large numbers for H, I , and T. Oh yeah, in order for there
to be that second carry, H has to be 9 – duh.
* Ok, H = 9, so then
I = 0, and T = 8. So far we have...
89REE
+ ONE
90PPO
* Let's go back to
trying values for E... E can't be 9 or 8 because H & T are 9 and
8.
* What if E equals
7? Then O = 4.
* What if N = 6,
then P would be 4 – but O is already 4, so NOT CORRECT
* What if N = 5,
then P would be 3, then R would be 8 – but T is already 8, so NOT
CORRECT
* What if N = 3,
then P would be 1, then R would be 6... I think that works! :-)
So T = 8, H = 9, R =
6, E = 7, O = 4, N = 3, P = 1, I = 0
89677 + 437 = 90114
YES!
Problem 4 – Which
is bigger, cube root of 24 or the fraction 10/3?
* Well, the cube
root of 24 is going to be a little bit under 3, because 3 cubed is
27. And 10/3 is over 3 because 9/3 would be 3. So therefore 10/3 >
(24)^(1/3)
* Or we could think
about getting fractions with common denominators. To make the cube
root of 24 into a fraction with three on the bottom, we need to
multiply the top and the bottom by 3. So then we can just compare
the tops of the fractions – which is bigger, 3*cube root of 24 or
10? The cube root of 24 is less than three so 3 times less than
three will give us less than nine. And ten is bigger than nine, so
again therefore 10/3 > (24)^(1/3)
* Or if we don't
like having a silly cube root, then we can cube both of the numbers
(since neither one is negative, we'll be ok in keeping the same
number bigger even after we cube). So now, which is bigger, 24 or
(10/3)^3? Well, (10/3)^3 is 1000/9, which is kinda a yucky fraction
still. We could do some quick long division, and get about 111,
which is a lot bigger than 24, and hence again therefore 10/3 >
(24)^(1/3)
Problem 5 – What
is two?
We often think of
two as referring to two of similar objects. When we're first having
kids learn to count, we place two crackers in front of them, and say
“one, two... two crackers”. But two can apply to all sorts of
different objects – and the two don't have to be similar. We can
have 'two things', which are not similar, except in their thingness.
And children have to learn that the two objects are still two, even
if we move them around or change the objects to different objects.
Two is also the
number that comes after one, and the number that comes before three.
Some children first learn the rote counting without the actual
correspondence to physical objects, and for them, two is just the
second syllable in the verbal string that they sing when asked to
count to ten... 'onetwothreefourfivesixseveneightnineten'.
Two can also apply
to two groups of objects, such as when kids start to learn
multiplication. Here, the total is much more than two objects, but
we can divide the objects into two groups – and we could also start
a discussion about evenness. Or a discussion of fractions... 4/2 =
2, 8/4 = 2, etc.
Two can also be a
measurement - a counting off along a continuous dimension, rather
than a counting of two discrete objects. Then we start thinking of
where two fits in with decimals – it's not just bigger than 1, it's
bigger than 1.9.
Two can also be a
numeral. In base ten, it can represent 2*10 in 320
Problem 6 – Part A
- One of your students can't understand why division by zero is
undefined. He thinks 0 divided by 0 should equal 0, explaining that
"if you have nothing and you divide it by nothing, then why
don't you end up with nothing?"
* But maybe 0
divided by 0 should equal 1, because don't we 'know' that any number
divided by itself is one?
* Or what if we let
0/0 = x, then if we move things around, we get 0 * x = 0. So x could
be 3 or 4 or 10 million or pi – we could put any number in and zero
times the number would equal zero. So therefore 0 divided by 0
equals all numbers?
* Or what if we
think of division as repeated subtraction? Eight divided by two can
be thought as how many times do you need to subtract two from eight
to get to nothing. So how many times do you need to subtract zero
from zero to get to nothing? True, zero times works, but so does
once, twice, and any number of times. So again we get 0/0 equals any
number?
So we call 0/0
undefined and indeterminate because we can't determine the answer.
Problem 6 – Part B
- Another of your students says that not only
is zero an additive identity (i.e. that zero plus any number gives
you back that number), but that zero is also an identity with respect
to subtraction.
* The student is
correct in that zero subtracted from any number results in the same
number, just as in the additive identity where zero added to any
number results in the same number. Although we don't usually talk
about a subtractive identity. Perhaps because subtraction can be
thought of as addition of a negative number? So subtracting zero is
like adding a negative zero? But what does negative zero mean? Or
maybe we don't talk about a subtractive identity because the
subtraction is not commutative? A + B = B + A, but A – B doesn't
equal B – A. So A + 0 = 0 + A, but A – 0 = A while 0 – A = -A.
Interestingly, some
of the early math research does refer to A + 0 = A as the subtractive
identity...
Problem 7 –
Reaction to article...
http://www.thephora.net/forum/archive/index.php/t-61001.html
Coming from a
cognitive science/education/language background, I knew most of the
experiments mentioned in the article, although hadn't heard of that
particular culture. The log-linear aspect of learning is actually
very important in early math, and seems to be a predictor of math
ability - “Moreover, those children who were able to correctly
estimate numbers' positions tended to score higher on the math
portion of the Stanford Achievement Test, Ninth Edition.”
http://www.apa.org/monitor/nov05/linear.aspx
Research shows that
playing board games can help to move young children from log to
linear...
And it works better
if the teacher/parent helps the child to use 'counting on'
language...
I was away on a
camping vacation right before the fall term started – away from
internet & technology, away from society, away from numbers –
at least in some ways. Like the indigenous people, I told time by
the sun – if a friend asked what time it was, I said morning,
midday, afternoon, evening, or night. I didn't do much counting, but
I also didn't feel like I was being 'productive'. Obviously we need
numbers to function in the society that we have now – numbers are
needed to create the words that I am typing on this screen, although
in some sense, it's all zeros and ones, so maybe we don't need the
higher numbers? :-)
No comments:
Post a Comment