Going to church doesn't make a presidential candidate a Christian any more than standing in a garage makes me an automobile. So if you think things can't get any worse, it's probably only because you lack sufficient imagination.
Teh Meh:
7-11
$3.16 as one of the numbers!
Well done, that was the hardest part.
"21"
Half right! :wink:
TTH:
I got the other 3 to be $1.20, $1.25 & $1.50
"Can you explain the answer so I can understand it?"
Sure thing doll; there's only one possible solution, namely:
7.11 = 3.16 + 1.25 + 1.50 + 1.20 = 3.16 x 1.25 x 1.50 x 1.20
Proof :
If a, b, c, d are the prices of the items expressed in (whole) cents, what we are told is that a+b+c+d = 711 and abcd = 711000000 = 2^6 3^2 5^6 79.
So, one (and only one) of the amounts, say a, is a multiple of 79. This is, of course, less than 9 times 79 (which is 711), and may thus only be 1,2,3,4,5,6, or 8 times 79 (7 times 79 is ruled out, because it does not divide 711000000). These 7 possibilities translate into the following equations:
1. a=79, b+c+d=632, bcd=9000000
2. a=158, b+c+d=553, bcd=4500000
3. a=237, b+c+d=474, bcd=3000000
4. a=316, b+c+d=395, bcd=2250000
5. a=395, b+c+d=316, bcd=1800000
6. a=474, b+c+d=237, bcd=1500000
7. a=632, b+c+d=79, bcd=1125000
Now, the product of 3 positive numbers of given sum is greatest when they are all equal, which means that the product bcd cannot exceed (b+c+d)^3/27. This rules out the last three of the above 7 cases.
In the first three cases, on the other hand, the sum b+c+d isn't a multiple of 5, so at least one of b,c,d (say d) isn't either. Therefore, the product bc must be a multiple of the sixth power of 5. Since neither b nor c can be large enough to be a multiple of the fourth power [625 is clearly too big a share of 711, leaving only 7 cents for two items in the first case and nothing at all in the other two] we must conclude that both b and c are nonzrero multiples of 125. The number d would thus be obtained by subtracting from one of three possible sums (632, 553, 474) some multiple of 125 (necessarily: 250, 375 or [in the first two cases] 500). This gives only 8 possible values for d (382, 257, 132, 303, 178, 53, 224, 99). Each is unacceptable because it has a prime factor other than 2 or 3.
Therefore, only the fourth case is not ruled out, so that a=$3.16.
a=316, b+c+d=395, bcd=2250000.
Since the sum b+c+d is a multiple of 5, and at least one of the terms is a multiple of 5, either only one is, or all are. The former option is ruled out since this would imply for the single multiple of 5 to be a multiple of 5^6=15625, which would, by itself, be much larger than the entire sum of 395. So, b, c, and d are all multiples of 5, and we may let b=5b', c=5c', d=5d', where b'+c'+d'=79 and b'c'd'=18000.
Now, these three new variables may not be all divisible by 5 (otherwise their sum would be too). It's not possible either to have a single one of them divisible by 5, because it would then have to be a multiple of 5^3=125 and exceed the whole sum of 79.
Therefore, we must have a multiple of 25 (say b'=25b") and a multiple of 5 (say c'=5c"): 25b"+5c"+d'=79 and b"c"d'=144. It is then clear that b" can only be 1 or 2. If b" was 2, then we would have 5c"+d'=29 and c"d'=72, implying that c" is a solution of x(29-5x)=72 or 5x^2-29x+72=0. However, this quadratic equation does not have any real solutions, because its discriminant is negative. Therefore, b" is equal to 1 and b=5(25b")=125=$1.25.
The whole thing thus boils down to solving 5c"+d'=54 and c"d'=144, which means that c" is solution of x(54-5x)=144 or 5x^2-54x+144=0. Of the two solutions of this quadratic equation (x=6 and x=4.8), we may only keep the one which is an integer. Therefore c"=6, c'=30, c=150=$1.50.
Finally, d'=144/c=24 which means d=5d'=120=$1.20.
Therefore, the solution is unique (the order of the 4 items being irrelevant): $3.16, $1.25, $1.50, $1.20.
Slippy:
FLAGS
Permute me to suggest: P(7,2) = 7! / (7-2)!
ie. 7x6
"Permute" That is one heck of an expensive education you have there, I am pleased to see it is not going to waste.
For those of us from the other side of the tracks; there are seven choices for the flag on top. There are six remaining choices for the flag below.
The Counting Principle says that the total number of ways is the number of choices for choice 1 times the number of choices for choice 2.
7 ways x 6 ways = 42 ways
Works every time!
A mother (please don't assume I am referring to Miss mi) and son have the same birthday; October 20. On that day the mother becomes 42 and her son becomes 11.
In what year will the mother be exactly twice as old as the son
29 = D in F in a L Y
32 = D F at which W F
40 = D and N of the G F
; but I know the answer to these:
I don't see 3 green words 
