Mark:
BEER
4
96 Beers
16 Sixers
Or 4 Cases
DO YOU KNOW
1. a
2. b
3. c
4. c
5. b
1. a-Dopey.
2. b-John Glenn.
3. c-Nathaniel Hawthorne.
4. c-Ian Fleming.
5. b-The Great Pyramids.
SEVEN ELEVEN
$1.20, $1.25, $1.50, $3.16
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 = 26 32 56 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 56=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 53=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 5x2-29x+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 5x2-54x+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). Well done Mark
Look at this sentence and see if you can find the reason why all the 2-letter words have a common trait. What is this trait
"Hi! Ma and Pa told me I'd better say that all the two-letter words in this sentence have something in common ..... or else!"
While Barb was out golfing with her husband George, they met Ron. Ron was Barb's only sister's mother-in-law's only son's grandfather in-law. What was Ron's relationship to Barb
A man left home one day and made three left turns and met a man with a mask on. What was the first man's profession
What number comes next in the following sequence
1/4, 1/2, 3/4, 1, 5/4, 3/2,
A quick math question: If all the numbers in existence starting with zero were written in English and then arranged in alphabetical order, what number would be first
The number 2002 reads the same backwards and forwards. (This is called a palindromic number). Name a palindromic number with an even number of digits
Welcome Nitin, are you related to Tintin? He was so good they named him twice.
You wrote, "Question: What happens after death?"
"Answer : Nobody lived to answer that.. "
I'm a nobody, and lived to tell the tale.
CYCLISTS
4
