A friend sent this internet address to check out, see what you think.
http://www.milaadesign.com/wizardy.html
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As Trevor Fougere so elequently summed up on the mail list...
The proof is fairly simple.
Any two digit number can be expressed as (10X + Y) where X, Y are elements of the set (0-9)
adding the two digits together gives you (X + Y)
so (10X + Y) - ( X + Y) = 10X + Y - X - Y
which = 10X - X + Y - Y
which = 10X - X + 0
which = 9X + 0
which = 9X
hence all are divisible by 9
Inerestingly is works for three digit numbers as well... briefly
100X + 10Y + Z - X - Y - Z = 99X + 9Y = 9 ( 11X + Y) = hence divisible by 9
Oh
and it works for 4 digit numbers ...5 digit numbers...
The proofs remain elemental as previously described.
I leave the proof for all positive integers greater than or equal to 10 to the reader.
The proof is fairly simple.
Any two digit number can be expressed as (10X + Y) where X, Y are elements of the set (0-9)
adding the two digits together gives you (X + Y)
so (10X + Y) - ( X + Y) = 10X + Y - X - Y
which = 10X - X + Y - Y
which = 10X - X + 0
which = 9X + 0
which = 9X
hence all are divisible by 9
Inerestingly is works for three digit numbers as well... briefly
100X + 10Y + Z - X - Y - Z = 99X + 9Y = 9 ( 11X + Y) = hence divisible by 9
Oh
and it works for 4 digit numbers ...5 digit numbers...
The proofs remain elemental as previously described.
I leave the proof for all positive integers greater than or equal to 10 to the reader.
Garth...
I'm having flashbacks to Calculus class. Don't you know "the reader" never does additional work recommended by the text?
Here's one for you if you are in a thinking mood:
A race car is lapping a one-mile circular track. He completes the first lap at an average speed of 30 MPH. How fast does he need to go on his second lap to average 60 MPH for both laps?
Difficulty: no pencil & paper. do it in your head.
I'm having flashbacks to Calculus class. Don't you know "the reader" never does additional work recommended by the text?
Here's one for you if you are in a thinking mood:
A race car is lapping a one-mile circular track. He completes the first lap at an average speed of 30 MPH. How fast does he need to go on his second lap to average 60 MPH for both laps?
Difficulty: no pencil & paper. do it in your head.
Too late, he used all his time up on the first lap.
You see, his first lap of the mile course at 30mph took two minutes. Now, he is challenged to do one more lap, giving a total of two miles, for an average of 60 mph. But two miles at 60 mph takes two minutes, and he has already spent that much time on the first lap.
Can't be done.
Can't be done.
Exactly. This one usually catches a few people. I guess Pantera owners are smarter than most.
quote:Exactly. This one usually catches a few people. I guess Pantera owners are smarter than most.
Cool! I guess I didn't get back from taking the kids out trick-or-treating early enough to try it myself.
Here's an appropriate one for you today...
Why do computer geeks celebrate Halloween on Christmas? Because Oct 31 is equal to Dec 25. Get it?
quote:Why do computer geeks celebrate Halloween on Christmas? Because Oct 31 is equal to Dec 25. Get it?
Not until I Googled it.
I figured it was something doing with Binary, but I'd never heard of Octal.
Thanks Garth, I'm a bit smarter this morning already.
Larry
He would have to go 186,120 miles per second.
quote:He would have to go 186,120 miles per second.
Even at the speed of light, he would be over two minutes and thus not quite at an average of 60 mph.
You noticed my figure exceeds the speed of light, so he would travel back ward in time and make up for his sluggish first lap.
quote:so he would travel back ward in time and make up for his sluggish first lap.
ROFLOL
the problem given was::
A race car is lapping a one-mile circular track. He completes the first lap at an average speed of 30 MPH. How fast does he need to go on his second lap to average 60 MPH for both laps?
____________________________________
ok, i admit that sometimes i am not the sharpest knife in the drawer, but.....
seems to me that this is a simple algebra problem (30mph + Xmph)/2 = 60mph with the solution of 90mph.
now, everyone else has been very emphatic in their answers, so what have i done wrong?????????
nazgul
A race car is lapping a one-mile circular track. He completes the first lap at an average speed of 30 MPH. How fast does he need to go on his second lap to average 60 MPH for both laps?
____________________________________
ok, i admit that sometimes i am not the sharpest knife in the drawer, but.....
seems to me that this is a simple algebra problem (30mph + Xmph)/2 = 60mph with the solution of 90mph.
now, everyone else has been very emphatic in their answers, so what have i done wrong?????????
nazgul
your time was consumed in the first lap.
Since average speed is always calculated as a distance (length) divided by a time, the units of average speed are always a distance unit divided by a time unit. Common units of speed are meters/second (abbreviated m/s), centimeters/second (cm/s), kilometers/hour (km/hr), miles/hour (mi/hr - try to avoid the common abbreviation mph)
George has planned a trip to the PI Motorsports holiday BBQ on Dec. 1 2007
http://www.pim.net/Holiday2007.pdf
----- mark your calendars guy’s -----
which is 60 miles away. He wishes to have an average speed of 60 miles/hour for the trip. Due to traffic, however, he only has an average speed of 30 miles/hour for the first 30 miles. How fast does he need to go for the remaining 30 miles so that his average speed is 60 miles/hour for the whole trip?
Most likely you thought "Oh, 90 miles/hour - since the average of 30 and 90 is 60! Boy, this is easy!"
Unfortunately, however, the answer is not 90 miles/hour. Here's why: You know that average velocity = distance/time (v = d/t). In order to have an average speed of 60 miles/hour over a distance of 60 miles, George must complete the trip in 1 hour:
60 MPH = 60 miles / 1 hour
But George has already taken an hour (it takes 1 hour to go 30 miles with an average speed of 30 miles/hour) - and he is only half way! It is impossible for him to complete the trip with an average speed of 60 miles/hour! He would have to install a 427 from PI Motorsports and go infinitely fast!
Speed = ∞
or
Speed = 427
Notice that it would take 1/3 of an hour to cover the last 30 miles at 90 miles/hour. The total time for his trip would be 1.33 hours, and his average speed would be:
Y = distance / time
or
60 miles / 1.33 hours = 45 miles per hour
- the average speed for the whole trip cannot ever be 60 miles/hour!
The moral of the story: Don't average averages!
George has planned a trip to the PI Motorsports holiday BBQ on Dec. 1 2007
http://www.pim.net/Holiday2007.pdf
----- mark your calendars guy’s -----
which is 60 miles away. He wishes to have an average speed of 60 miles/hour for the trip. Due to traffic, however, he only has an average speed of 30 miles/hour for the first 30 miles. How fast does he need to go for the remaining 30 miles so that his average speed is 60 miles/hour for the whole trip?
Most likely you thought "Oh, 90 miles/hour - since the average of 30 and 90 is 60! Boy, this is easy!"
Unfortunately, however, the answer is not 90 miles/hour. Here's why: You know that average velocity = distance/time (v = d/t). In order to have an average speed of 60 miles/hour over a distance of 60 miles, George must complete the trip in 1 hour:
60 MPH = 60 miles / 1 hour
But George has already taken an hour (it takes 1 hour to go 30 miles with an average speed of 30 miles/hour) - and he is only half way! It is impossible for him to complete the trip with an average speed of 60 miles/hour! He would have to install a 427 from PI Motorsports and go infinitely fast!
Speed = ∞
or
Speed = 427
Notice that it would take 1/3 of an hour to cover the last 30 miles at 90 miles/hour. The total time for his trip would be 1.33 hours, and his average speed would be:
Y = distance / time
or
60 miles / 1.33 hours = 45 miles per hour
- the average speed for the whole trip cannot ever be 60 miles/hour!
The moral of the story: Don't average averages!
Great way to explain it!