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#1 Re: Puzzles and Games » Find the numbers! (Algerbra problem) :) » 2008-02-04 17:24:06

Yep, I messed up on calculating G.   But I was still correct because I said one of the possibilities was that I did something wrong!   CYA!

#2 Re: Puzzles and Games » Find the numbers! (Algerbra problem) :) » 2008-02-04 11:35:21

DONALD
GERALD
=====
ROBERT

T=0, Therefore D must be 5

5ONAL5
GERAL5
=====
ROBER0

From O + E = O, E must be 0 or 9 (carry one).   Can't be 0, must be 9

5ONAL5
G9RAL5
=====
ROB9R0

From A+A=9, A must be either 4 or 9 with a "carry one" from L+L.  Can't be 9.

5ON4L5
G9R4L5
=====
ROB9R0

So far we have:
T = 0
D = 5
E = 9
A = 4

We still have 1, 2, 3, 6, 7, 8.   Using L + L = R (plus a "carry one") and R + G = 5 (also a carry one) we have the following possibilities

L =  1   2   3   6   7  or 8
R =  3   5   7   3   5  or 7
G =  1  -1 -3   1  -1 or -3

So G=1, R=3 and L=6

5ON465
193465
=====
3OB930

That leaves 2, 7 and 8 for B, N and O.   We know N+3=B but none of the number combinations we have left will work.   So either:  I did something wrong, I assumed something I shouldn't have (there is a 1-to-1 mapping between letters and numbers), the problem as stated doesn't have a valid answer.

#3 Re: Help Me ! » binary numbers » 2008-02-01 16:34:48

This may be the method you're referring to.    Convert 70 (decimal) to binary.

70 / 2 = 35   R0
35/2   = 17   R1
17/2   = 8     R1
8/2     = 4     R0
4/4     = 2     R0
2/2     = 1     R0
1/2     = 0     R1

You answer is the remainders reading upwards:  1000110

Works for converting to any number system:  949 to base 16:

949 /16 = 59    r5
59 / 16 = 3      r11 (hex B)
3 / 16   = 0      r3

949 (decimal) = 3B5

#4 Re: Help Me ! » probability » 2008-01-21 16:48:36

635,013,559,600 / 4 = 158,753,389,900 .

That's (52 choose 13) divided 4. 

Did that happen to you EMPhillips?   If it ever does, just realize that with you having all of the cards in one suit, the chances of one of the other 3 players  having all 13 cards of one of the other suits drops to a mere 1 in 2,707,475,148!

#5 Re: Help Me ! » Who can solve this problem ? I can . How about you????? :) » 2008-01-18 04:49:02

x = -3

(x+2) * (X+2) = 1
(x+3)(x+3)(X+3) = 0
(X+4)(X+4)(X+4)(X+4) = 1

1+0+1 = 2

#6 Re: Coder's Corner » two's complement number system » 2008-01-17 03:10:40

It simplifies adding a positive number to a negative number.

-4          1100         
+5          0101
==         ====
  1         10001


-5          1011
+4          0100
==        ====
-1          1111

If there are 5 digits (in these examples) in the answer, the leftmost digit is ignored.

#7 Re: Coder's Corner » two's complement number system » 2008-01-16 17:44:18

The first bit (leftmost) is the sign bit and is not used in computing the value of the number.   Zero signifies a positive number, a one signifies a negative.   So with 4 bits, you can represent from 7 (0111) to -8 (1000).

#8 Re: Help Me ! » What's this for? » 2007-12-30 17:33:15

n(n+1)/2 is the summation of 1 through n.   That's the closest thing I can think of.

#9 Re: Puzzles and Games » Lateral thinking puzzles » 2007-12-25 15:21:47

I like Jane's answer to 16 better but the real reason is:

#10 Re: Coder's Corner » MNEMONICS. » 2007-12-14 15:38:10

HOMES - Huron, Ontario, Michigan, Erie and Superior - the 5 Great Lakes

Sally can tell Oscar has a hat on always (reading down)

Sine (sally) = Opposite (oscar) / hypotenuse (has)
Cosine (can) = Adjacent (a) / hypotenuse (hat)
Tangent (tell) = Opposite (on) / adjacent (always)

If you want to make it even more likely that kids will remember, change "hat" to "hard"!

#12 Re: Help Me ! » a shopkeeper » 2007-12-11 04:13:49

He is going to use the scale that weighs on the light side to buy goods.   Let's say the goods really weight 10 pounds.   He's going to pay for only 9 pounds but he's going to sell them as 11 pounds.   That's a 2 pound difference or 22% gain (2/9).

#13 Re: Ganesh's Puzzles » Oral puzzles » 2007-12-07 17:09:35

Quote from JohnnyReinB: "What happens if you raise a number to a fraction?"

x ^ (1/2) is the square root of X
x ^ (1/3) is the cube root of X
.
.
.

#14 Re: Guestbook » fhh » 2007-12-04 09:13:06

You're correct Jonathon.

#15 Re: Help Me ! » Help with finding the percentage » 2007-12-01 17:18:00

The way I was taught:

Later on, you will realize that "of" (as in 25% OF 25000) implies multiplication.   And 25% = .25, so:

.25 * 25000 = 6250

#16 Re: Help Me ! » Are there any calculators that can convert bases up to Hexadecimal? » 2007-11-14 16:43:31

certainly.   There are hand-held calculators, software calculators* and there is even one on mathisfun (http://www.mathsisfun.com/numbers/convert-base.php).

*If you're using some version of Windows, go to Start and click on Run.  Type in calc (short for calculator) and hit return.  From the View menu item, make sure "scientific" is selected.   That will give you a calculator with hexadecimal, octal and binary capabilities in addition to the standard decimal system.

#17 Re: Help Me ! » Find the number succeeding and preceeding EOE base 12 » 2007-11-14 02:34:10

1 to 36 in base 12:

  1      2     3    4     5     6    7     8     9   A     B   10    (1 through 12)
11    12   13   14   15   16   17   18   19  1A   1B   20    (13 through 24)
21    22   23   24   25   26   27   28   29  2A   2B   30    (25 throught 36)

I think MathIsFun is probably right on his speculation.   Can you give us the exact question and instructions as it was given to you?

#18 Re: Help Me ! » Find the number succeeding and preceeding EOE base 12 » 2007-11-13 16:36:14

You're not the only one confused.   Base 12 would use the digits 0-9 plus A and B.  "E" would not be a valid digit.   And is that an zero or the letter O?

#19 Re: Help Me ! » probability » 2007-11-11 16:18:25

Let X be the number of white balls in the 2nd urn.

The chances of getting 2 white balls is 6/10 * x/(16+x).
The chances of getting 2 red balls is 4/10 * 16/(16+x).

Add the 2 together and they are equal to the .44 probability you were given.

(6x + 64) / (160 + 10x) = .44
6x + 64 = 4.4x + 70.4
1.6x = 6.4
x = 4

#23 Re: Help Me ! » Algebra » 2007-10-22 16:25:06

Whoops.   Sorry about that Daniel.   And thanks, Mathsyperson.  I was a little too hasty.   If I solved it all the way through and then double-checked, I would have seen I was the one who made the mistake

I'll do now what I should have done before.  Continuing with what you already figured out....
p = -12/7  and q = 3p - 18

q = 3(-12/7) - 18
q = -36/7 - 126/7 = -162/7

Plug p and q back into your original question to double-check.

f(x) = 2x² - 12x/7 -162/7
f(-3) = 2(9) + 36/7 - 162/7
        = 18 - 126/7
        = 18 - 18 = 0

f(4) = 2(16) - 48/7 - 162/7
      = 32 - 210/7
      = 32 - 30 = 2

Much better!

#24 Re: Help Me ! » Algebra » 2007-10-22 10:09:04

Edited by Pi Man:  Ignore this post (#2) .   I made the mistake, not Daniel.   

Daniel123 wrote:

f(4) = 2(4)² + 4p + q - 2 = 0
      = 30 + 4p + q = 0

f(4) = 2(4)² + 4p + q - 2 = 0
      = 32 + 4p + q = 0                   32, not 30
q = -4p -32

-4p-32 = 3p - 18
-14 = 7p
p = -2

#25 Re: Help Me ! » probability question » 2007-10-16 02:51:57

Sounds like you're playing some Texas Hold'em!

You have 2 cards in your hand, leaving 50 cards that the other 5 will come from.   There are 2118760 different sets of 5 cards out of those 50.   (That's "50 choose 5" = 50! / (5! * 45*)).

There are 11 spades left in the deck.   The number of ways exactly 3 spades can show up in the next 5 cards is:

(11 choose 3) * (39 choose 2)            ! you want 3 of the the remaining spades plus 2 non-spades

(11! / (8! * 3!)) * (39! / (37! * 2!))
165 * 741 = 122265

So the odds of exactly 3 spades being flopped is the number of "good" hands divided by the total number of possible hands:  122265/2118760 =~ 5.7%

If you want the odds of 3 or more spades, you need to calculate the the number of possible hands with 4 spades and 5 spades:

4 spades = (11 choose 4) * (39 choose 1) =  330 * 39 = 12870
5 spades = (11 chosoe 5) * (39 choose 0) =  462 * 1 = 462

(122265 + 12870 + 462) / 2118760 =~ 6.4%


For the example you provided, it's really more complicated than that.   The odds of getting making a pair is different than the odds of making a pair or better (2 pair, 3 of a kind, full house, etc.).   I'm sure the odds of all these events are already out there on the internet.   If you interested in how to compute them, you've come to the right place!

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