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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 is not a multiple of 5, so at least one of b,c,d (say d) is not either. This means that 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 multiples of 125 (the cube of 5).
Therefore, only the fourth case can be valid, 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 these three is a multiple of 5, either only one is, or all three are. The former possibility 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, it's not possible that all three of these new variables are divisible by 5 (otherwise their sum would be). 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.
As the order of the items is irrelevant, the problem has therefore a unique solution and the only possible costs of the 4 items are (in the order "discovered" above):
$3.16, $1.25, $1.50, $1.20.
Special thanks to Numercicana.com |
