Rsaturn9

Excellent Math Trick, Thanks Stevonator65:approve:

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stevenator65

Originally Posted by kor

How about 9^9^9 andy

Winner Winner Winner

Here's the next one:

You must read it one line at a time. Do not skip ahead.

1) pick a number from 1-9


2) subtract 5

Don't look ahead, one line at a time!
3) multiply by 3


4) square the number (multiply by the same number -- not square root; and yes, you can square 0. It equals 0)


5) add the digits until you get only one digit (i.e. 64=6+4=10= 1+0=1)


6) if the number is less than 5, add five. Otherwise subtract 4.


7) multiply by 2


8) subtract 6


9) map the digit to a letter in the alphabet 1=A, 2=B, 3=C, etc...


10) pick a name of a European country that begins with that letter


11) take the second letter in the country name and think of a mammal that begins with that letter


12) think of the color of that mammal




You have a gray elephant from Denmark.

soulmirror

Originally Posted by stevenator65

Winner Winner Winner

See, now the problem with that sort of puzzle is that there's no consistency. It's true that

9^9^9 > 999

However, I'm pretty sure that

999! > 9^9^9

Think about it...9^9^9 = 387420489^9 = 387420489*387420489*387420489*...*387420489

But if I consider a sequence of multiplications in 999!, say

899*898*897*896*895*894*893*892*891 = 3.6x10^26

Then there are more than eight other sequences larger than this one -- such as the next nine numbers up, the next nine numbers after those, and so on. So if I multiply just those nine sequences together, I get a number much, much larger than 9^9^9 already. And there are far more numbers to multiple to get to 999!

So 999! is, unless I am really mistaken, a much, much larger number than 9^9^9. But why doesn't it count?

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stevenator65

Soulmirror, 999! has 4 digits. But good try!

Here's another:

1.Have someone pick a number between 1 and 9.
2.Now have him use a calculator to first multiply it by 9, and then multiply it by 12,345,679 (notice there is no 8 in that number!).
3.Have the person show you the result so you can tell him the original number he selected!
How? Easy. If he selected 5, the final answer is 555,555,555. If he selected 3, the final answer is 333,333,333.

The reason: 9 x 12345679 = 111111111. You multiplied your digit by 111111111. (By the way, that 8-digit number (12,345,679) is easily memorized: only the 8 is missing from the sequence.)

soulmirror

Originally Posted by stevenator65

Soulmirror, 999! has 4 digits. But good try!

Why is a ^ not a digit but a ! is a digit? They're both arithmetic operators! By your logic, 9^9^9 has five digits.

Well, fun to play anyway.

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stevenator65

if you write 9 to the power of 9 to the power of 9 it's a 9, then a nine slightly above and to the right, then a 3rd 9 slightly above and to the right. There are no "^" marks in it. We only use those on computers.

OklahomaMama

my brain hurts

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yeah, I'm still around :D

uzziah0

it is slight, as I would not say ! is a digit, but an operator;
and you can write 9^9^9 as

I'd say we're on a fine line, and both are good answers.
I was impressed with the 999!

Empel1960

Correct me if I'm wrong but 9^99 is larger than 9^9^9.

9^99= 2.9512X10^94
9^9^9=1.9662X10^77

stevenator65

Empel, my source says you're wrong. Good try.

Here's the next one:
Three people check into a hotel. They pay $30 to the manager and go to their room. The manager suddenly remembers that the room rate is $25 and gives $5 to the bellboy to return to the people. On the way to the room the bellboy reasons that $5 would be difficult to share among three people so he pockets $2 and gives $1 to each person. Now each person paid $10 and got back $1. So they paid $9 each, totalling $27. The bellboy has $2, totalling $29. Where is the missing $1?

uzziah0

10 + 10 + 10 = 30

30 - 5 = 25
5 - 2 = 3

the three goes back to the original 10s
so that becomes 9 + 9 + 9 = 27
27 - 2 = 25 (don't add the 2, subtract it).

as for 9^9^9, it depends on the order of operations:
(9^9)^9 is not the same as 9^(9^9)

atompson

Originally Posted by Empel1960

Correct me if I'm wrong but 9^99 is larger than 9^9^9. 9^99= 2.9512X10^94 9^9^9=1.9662X10^77

You are correct.

stevenator65

I stand corrected. I appreciate the input. Did anyone answer the bellboy one?

uzziah0

Originally Posted by stevenator65

I stand corrected. I appreciate the input. Did anyone answer the bellboy one?

I did.
10 + 10 + 10 = 30

30 - 5 = 25
5 - 2 = 3

the three goes back to the original 10s
so that becomes 9 + 9 + 9 = 27
27 - 2 = 25 (don't add the 2, subtract it).

stevenator65

Originally Posted by uzziah0

I did. 10 + 10 + 10 = 30 30 - 5 = 25 5 - 2 = 3 the three goes back to the original 10s so that becomes 9 + 9 + 9 = 27 27 - 2 = 25 (don't add the 2, subtract it).

We have a winner!!!!

Here's the next one:

If I were in Florida and dropped a baseball into a bucket of water which was at a temperature of 45 degrees F and dropped another identical baseball into a identical bucket of water at a temperature of 30 degrees F, which baseball would hit the bottom of the bucket first?