helllllppppppppp

haiying · 2011-03-27 20:10:37

1. The following is the actual solution of a second grader. Describe the algorithm used. Determine whether or not it is a valid algorithm. If not, state where it fails. If so, justify the algorithm.

Question: How much is 42 divided by 7?
Solution: 40 divided by 10 is 4;
3 and 3 and 3 and 3 is 12;
12 plus 2 is 14;
14 divided by 7 is 2;
2 plus 4 is 6.
The answer is 6.

bobbym · 2011-03-27 21:06:48

Hi haiying;

His/her algorithm for a / b is shown below on the left.

As you can see it reduces to a / b so I expect it to be correct for all fractions ( division problems ).

There are only 2 possibilities:

1) He/she is brilliant!
2) He/she, cheated and looked it up on the internet and it is some method I have never seen before.

In either case this person will make a fine mathematician/programmer and better yet, a great problem solver. Is this person interested in pursuing a mathematical life? Any chance you can direct him/her to this forum?

John E. Franklin · 2011-03-29 06:24:03

@bobbym: if you have the time, could you give me a hint as to
how to simplify your cool mod equation (I see it works great for many times table problems!!)
so it equals a/b. You see, I never have studied "mod" algebra simplification yet.
What do you call "mod" math also?

bobbym · 2011-03-29 09:55:32

Hi John E. Franklin;

How are you? Sorry for the delay but I had to go shopping for furniture.

your cool mod equation

Actually, his second grader invented or discovered that equation. He just phrased it differently.

What do you call "mod" math also?

Congruences is what it is called and it was discovered by Gauss.

@bobbym: if you have the time, could you give me a hint

I would love to John, but Mathematica did that one and I might not be smart enough to duplicate it, but I will try.

Okay, I got lucky! Make the substitution v = a mod 10. Then you get:

Add the two fractions on the LHS:

Expand out the first term on top of the LHS.

The b(a-v) and -b(a-4) cancel.

Expand the first term on top of the LHS and clean up.

10v and -10v cancel.

Cancel the 10 in the top and bottom.

John E. Franklin · 2011-03-30 02:53:45

I'm doing alright, thanks for asking, though last October my Dad died and a close friend of only 40 years passed too.
I never got her on this math forum, but she had a masters in mathematics. It is very unfortunate she passed on.
She was on meds all her life, and that contributed to a major weight problem about twice that of mine.

But back to this problem here. Thanks for showing me you can substitute v for the mod stuff. That is awesome to know.
I showed this 2nd graders technique to my Mom, and she loved it. She also noticed that between 45 and 81, given you
are picking numbers from the times tables studied in 2nd or 3rd grade, you can do a kludge that is faster, but it doesn't
work below 45. As an example, it is easier for me, 56/7, do the 7mod(10) and get 3 or do 10-7 for 3. Then add
the 3 to the 5 in 56 and get the answer of 8. Now this works for about ten problems between 45 and 81, for what it's
worth. As another example: 49/7: 10-7=3 4 from 49 plus 3 is 7, the answer. But it is a kludge; works by
reason of being within a gap region of coincidence.

bobbym · 2011-03-30 03:01:50

Hi John;

Sorry to hear about your Dad and your friend.

Glad your Mom liked it, I do not know where that 2nd grader got his idea from. Just that it is very clever. The OP has never returned to tell us more.

John E. Franklin · 2011-03-30 03:24:11

It would be very beneficial if we could find or develop
this formula or method so that it works for 3 and 4 digit
numbers as well. I did an an example with 3 digits, and
treated the first 2 digits as one digits, and I used subtraction
from 10 instead of mod. and I multiplied negative numbers
together on paper, and it all worked out correctly, but it
was not a time saver by any means!!

bobbym · 2011-03-30 05:09:22

Hi John;

It already works for 3 digits divided by 2 digits. To avoid the negative numbers change every 10 to a 100.

John E. Franklin · 2011-03-30 09:44:01

Let me try, okay how about 210/14, a known integral answer...
21 x (100-14) = 21 X 86 = 1606 + 0080 + 0120 = 1606 + 0200 = 1806
1806 + 14 is 1820 and 1820 / 14 is 130 and 130 plus 21 is 151.
Hmmm where did I go wrong. the answer was supposed to be 15.
Any clues??

bobbym · 2011-03-30 10:06:14

Hi;

Comments are in parentheses.

a1=210: b1 = 14:

r1 = 210 mod 100 ( r1 = 10 )

a2 = a1 - r1 (a2 = 200)

r2 = 100 - b1 (r2 = 86 )

t = a2 / 100 ( t = 2)

a2 = t * r2 + r1 ( a2 = 182 )

a2 / b1 + t ( result = 15 )

As you can see this is not practical because the division in the last step is almost as large as the original problem. So maybe it is best to go back to his 10 rather than 100. That will produce smaller numbers.

Math Is Fun Forum

#1 2011-03-27 20:10:37

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#2 2011-03-27 21:06:48

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#3 2011-03-29 06:24:03

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#4 2011-03-29 09:55:32

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#5 2011-03-30 02:53:45

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