Puzzles with numbers and things

[snookersfun]

a short update on puzzle 72. I have 4 correct answers so far (well done d-g, abextra and the newcomers Sarmu and austrian_girl)

... now here is some fun:
THE TEN COMMANDMENTS OF MATHEMATICS
1. Thou shalt read Thy problem.

2. Whatsoever Thou doest to one side of ye equation, Do ye also to the other.

3. Thou must use Thy "Common Sense",
else Thou wilt have flagpoles 9,000 feet in height, yea ... even fathers younger than sons.

4. Thou shalt ignore the teachings of false prophets to do work in Thy head.

5. When Thou knowest not, Thou shalt look it up, and if Thy search still elude Thee, Then Thou shalt ask the all-knowing teacher.

6. Thou shalt master each step before putting Thy heavy foot down on the next.

7. Thy correct answer does not prove that Thou hast worked Thy problem correctly. This argument convincest none, least of all, Thy teacher.

8. Thou shalt first see that Thou hast copied Thy problem correctly before bearing false witness that the answer book lieth.

9. Thou shalt look back even unto Thy youth and remember Thy arithmetic.

10. Thou shalt learn, speak, write, and listen correctly in the language of mathematics, and verily A's and B's shall follow Thee even unto graduation.

[davis_greatest]

Ape Break Madness - update

And now snookersfun has matched Sarmu and found the highest possible break!

I've also had a very good submission from austrian_girl - high, but not the maximum possible

chasmmi has also submitted a very good break (after a couple of corrections along the way ) - it's high ... a bit higher than austrian_girl's, but not as high as the ones by Sarmu and snookersfun.

[davis_greatest]

Round 74 - Gordon in his prime

Don't forget that rounds 68, 72 and 73 are still open!

In the meantime, here is another...

Round 74 - Gordon in his prime

Charlie is teaching Gordon about prime numbers. "A prime number, Gordon," says Charlie, "is a positive whole number with exactly 2 different positive whole numbers as its factors: 1 and the prime number itself."

"Oh," says Gordon. "So is 1 prime?"

"No," replies Charlie. "Of course not. Because 1 only has one factor: 1 itself!"

"What about 15?" asks Gordon.

"Don't be silly," says Charlie. "15 is divisible by 1, 3, 5 and 15."

"I get it," says Gordon, "so the prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,..."

"Yes, yes!" interrupts Charlie. "Good little gorilla. Right, what are the prime factors of 110?"

"2, 5 and 11," says Gordon

"Good. And what are the prime factors of the number I get if I find the sum of 123456789 to the power of 123456789, added to 987654321 to the power of 987654321?"

"Um..." begins Gordon.

"It's OK," says Charlie. "Just tell me its 4 smallest, distinct prime factors."


Can you help little Gordon?

Answers (with explanation) may be posted on this thread.

[davis_greatest]

snookersfun is having a brave attempt at this but no one has got there yet! If there isn't much progress soon, I'll start posting hints.

[davis_greatest]

Scoreboard update

So I don't get confused with all the answers coming in by Private Message, I'll update the scoreboard now with points scored so far for the rounds that are still open...

Round 68 - Ape Break Madness

I am awarding

-: 1 point each to austrian_girl and to chasmmi for their high breaks (I won't say what they are yet, as the round is still open)
-: 2 points to sarmu for submitting a break which is also the highest possible
-: 1 point to snookersfun for submitting a break which is also the highest possible... she only gets 1 point because she has already scored more than 2 sets of double points on Ape Break rounds and also only gradually reached the maximum break.... it was also her suggestion that only new members should be eligible for double points!

Round is still open for anyone else who wants to try!



Round 72: Find the angle

snookersfun, please add points scored so far on your geometry round 72 to the table below.

Round is still open for anyone else who wants to try!


Round 73 - Add times bumble

This isn't easy - you'll soon see that you come up with 2 equations and 4 unknowns! But a bit of thought can cut down the number of possibilities quite drastically. Most correct submissions so far used a computer to search through every possibility!

Points awarded so far are:

-: 1 point to each of sarmu, elvaago and snookersfun

Round is still open for anyone else who wants to try!


SO HERE IS THE SCOREBOARD AFTER POINTS HAVE BEEN ADDED FOR SOLUTIONS RECEIVED SO FAR TO ROUNDS 68 AND 73, WHICH ARE STILL OPEN, BUT BEFORE ADDING POINTS FOR ROUND 72 (WHICH IS ALSO STILL OPEN, till Saturday noon GMT), AND BEFORE ROUND 74

snookersfun……………………….…..34½
abextra...............................16
davis_greatest.....................14½
Vidas..................................12½
elvaago...............................7
chasmmi..............................6½
robert602.............................6
The Statman……………………..……5
sarmu..................................3
Semih_Sayginer.....................2½
April Madness.......................1
austrian_girl.........................1

[snookersfun]

updated scoreboard for points scored in round 72 so far (d_g, a_g, abextra and sarmu)-round still open

SO HERE IS THE SCOREBOARD AFTER POINTS HAVE BEEN ADDED FOR SOLUTIONS RECEIVED SO FAR TO ROUNDS 68 AND 73, WHICH ARE STILL OPEN, AFTER ADDING POINTS FOR ROUND 72 (WHICH IS ALSO STILL OPEN, till Saturday noon GMT), AND BEFORE ROUND 74

snookersfun……………………….…..34½
abextra...............................17
davis_greatest.....................15½
Vidas..................................12½
elvaago...............................7
chasmmi..............................6½
robert602.............................6
The Statman……………………..……5
sarmu..................................4
Semih_Sayginer.....................2½
austrian_girl.........................2
April Madness.......................1

[davis_greatest]

Scoreboard update

abextra has also come in with a solution to Ape Break Madness - also the highest break possible! As she has scored double points on Ape Breaks before, she is also only eligible for 1 point.

SO HERE IS THE SCOREBOARD AFTER POINTS HAVE BEEN ADDED FOR SOLUTIONS RECEIVED SO FAR TO ROUNDS 68, 72 AND 73, WHICH ARE STILL OPEN, AND BEFORE ROUND 74

snookersfun……………………….…..34½
abextra...............................18
davis_greatest.....................15½
Vidas..................................12½
elvaago...............................7
chasmmi..............................6½
robert602.............................6
The Statman……………………..……5
sarmu..................................4
Semih_Sayginer.....................2½
austrian_girl.........................2
April Madness.......................1

[davis_greatest]

I will give at least half a point to anyone who finds the first 3 prime factors.

Finding the fourth is a little more difficult if you don't know where to start. For anyone who wants to make a serious attempt and learn a useful little theorem at the same time, this could help you!

http://en.wikipedia.org/wiki/Fermat%27s_little_theorem

Don't forget that rounds 68, 72 and 73 are still open!

In the meantime, here is another...

Round 74 - Gordon in his prime

Charlie is teaching Gordon about prime numbers. "A prime number, Gordon," says Charlie, "is a positive whole number with exactly 2 different positive whole numbers as its factors: 1 and the prime number itself."

"Oh," says Gordon. "So is 1 prime?"

"No," replies Charlie. "Of course not. Because 1 only has one factor: 1 itself!"

"What about 15?" asks Gordon.

"Don't be silly," says Charlie. "15 is divisible by 1, 3, 5 and 15."

"I get it," says Gordon, "so the prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,..."

"Yes, yes!" interrupts Charlie. "Good little gorilla. Right, what are the prime factors of 110?"

"2, 5 and 11," says Gordon

"Good. And what are the prime factors of the number I get if I find the sum of 123456789 to the power of 123456789, added to 987654321 to the power of 987654321?"

"Um..." begins Gordon.

"It's OK," says Charlie. "Just tell me its 4 smallest, distinct prime factors."


Can you help little Gordon?

Answers (with explanation) may be posted on this thread.

[abextra]

I would be happy with half a point for prime factors 2, 3 and 5.

[davis_greatest]

I am going to give half a point each to snookersfun and to abextra, who each found 2, 3 and 5.

Another half point is available for the 4th smallest prime factor of this very big number...

[Sarmu]

taking a guess 37?

[snookersfun]

taking a guess 37?

Sarmu, great choice there

Otherwise, the deadline to round 72 has passed now without further input. This round is now concluded as well.

[davis_greatest]

taking a guess 37?

37 is a factor, and snookersfun tried that too! But it is not the 4th smallest!

[davis_greatest]

I am going to give half a point each to snookersfun and to abextra, who each found 2, 3 and 5.

Another half point is available for the 4th smallest prime factor of this very big number...

We have found 2, 3 and 5 as factors.

2 is clearly a factor, since we are adding two odd numbers so must get an even number.

3 is clearly a factor, since 123456789 and 987654321 are both divisible by 3.

And 123456789 to the power of an odd number (123456789) must end in 9; while 987654321 to the power of any whole number must end in 1; so the sum must end in 0. This means that 5 is also a factor (it also shows that 2 is).


I will now tell you that 7 is the 4th smallest prime factor - half a point to anyone who can prove it.

In the meantime, half a point to each of snookersfun and abextra for giving the first 3 prime factors...

SO HERE IS THE SCOREBOARD AFTER POINTS HAVE BEEN ADDED FOR SOLUTIONS RECEIVED SO FAR TO ROUNDS 68, 73 and 74, WHICH ARE STILL OPEN

snookersfun……………………….…..35
abextra...............................18½
davis_greatest.....................15½
Vidas..................................12½
elvaago...............................7
chasmmi..............................6½
robert602.............................6
The Statman……………………...……5
Sarmu..................................4
Semih_Sayginer.....................2½
austrian_girl..........................2
April Madness........................1

[snookersfun]

Attached is a figure, containing a parallelogram ABCD. Also given is that the two angles on C are of same size and ED=AD.

Question: how big is the angle on F (with explanation of course)

Just noticed, I never gave the solutions, here are two possible ways of finding the angle on F, which turned out to be 9O degrees.

Austrian_girl's:
The angle on F is a right angle (90°).

Here comes my explanation. It may not be the simpliest way but anyway: Alle Wege führen nach Rom.
There are two identical angles on C, referred to as 2g (as I can't find a bloody gamma symbol ). This tells us that the angle on D (within parallelogramm) is 180-2g or 2(90-g). Consequently the angle on D (facing triangle) is 180-2(90-g) = 2(90-(90-g)) = 2(90-90+g) = 2g. Ok, this was the baby version! Anyway, I didn't want to leave out a step. Now we know that ED = AD. So both angles at A and E are of identical size. So the angle on E is (180-2g)/2 or simply 90-g. Now, considering the triangle ECF, we have two angles already defined, namely the angle on E (which is 90-g) and the angle on C (which is simply g). As we now, the sum of all angles within a triangle is 180°. So, 180-(90-g)-g = 180-90+g-g = 90°. q.e.d. (hopefully!!!)

Abextra’s:
The angle on F is 90'. As ABCD is a parallelogram, the angles on D and C are equal. As AD and ED are equal, the line that divides the angle on D makes 90' angle(s) with AE. The line is parallel with CF, so the angle on F is also 90'.