Math Challenge

Nov 2019
8
0
here
Hi, there.
For me 7 and 8 were always difficult to work with since school. If it is 9 for me it is 10-1, if it is 6 for me it is 5+1. Even now as a grown-up i tend to avoid these numbers in addition and multiplication.
I would probably do
20 + 40 8 + 7 and finally 60 + 15

But even in that way for me there is much to remember: 1st sum, 2nd sum and then to sum two sums. You wouldn't believe me but every time I have to ADD 8 and 7, I simply can't remember THAT sum. With the other sums I have no problem.
 

skipjack

Forum Staff
Dec 2006
21,181
2,332
You could do 27 + 48 = 27 + 50 - 2 = 77 - 2 = 75.
 
Nov 2019
8
0
here
Or we can do it like computers - mechanically (I am not sure about the name - ?complimentary numbers? I love these numbers because there's nothing to remember, just pure mechanics)
27= 100 - 73
48= 100 - 52
so: 27 + 48 = 100 - 73 +100 - 52 =100 + 100 -125 = 100 + 100 - 100 -25 = 100 - 25 (complimentary) = 75
Of course I would never do that.
Or
27 + 48 = 20 + 40 + 10 - 3 + 10 - 2(complimentary to 10) = 60 + 20 - 5 = 80 - 5 = 75

That trick goes "well" for 6, 7, 8, 9 because their complimentary numbers are below 5.

This reminded me about scary SUBTRACTION. I prefer not to subtract, but if I have to, I do this:

11'234 - 879 = 10'000 + 1'234 - 879 = 9'999 +1 +1'234 - 879 = 9'999 - 879 + 1'234 + 1 = 9'120 + 1'234 +1 = ...
Please don't ask about Division.

I think that teachers who teach children how to read, spell, add, subtract, multiply and divide are THE REAL HEROES.
 
Nov 2019
8
0
here
My bad, sorry

11'234 - 879 = 10'234 + 1'000 - 879(complimentary) = 10'234 + 121 =...
/999 - 879 + 1/
 

SDK

Sep 2016
689
463
USA
I often joke to my wife that $1+ 1 = 2$ is the hardest math equation I can't find a way to screw up. So for this I would write
\[27 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 \] and then \[48 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1\] so then its easy to see that
\[
27 + 48 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 75
\]
 
Jul 2008
5,205
47
Western Canada
20+40=60
7+8=15
60+15=75

Edit:
In fact, the above is a somewhat simplified summary of what went on in my brain. The actual process is as follows. Starting with:

27 + 48 = ?

Convert to internal brain data representation:

*&%TUGLU^EUY^765089u0387wrfy%R&^(&^@*^@%O*&%@*^#@*&*^%RIUYFI&^R(*&^RE*^%F*^%RFI&F)*&GF%@DF^%F*^%RT(&%G

Perform math operation:

5h08*^&*^#@ug^#@*&dup984gkp985#(*Rft6496&*^@&^%(&%@(&^@t984hq756i5)(*%*^yh9u

Convert result to normal math notation:

75
 
Last edited:
Dec 2019
1
0
Sydney
I have a problem that I hope someone can answer. If you had 10 couples walk to a continent who had a life span of 50 years and had 3 children per couple per generation. What would be the population after 50000 years ?