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Adding and Multiplying to Get 7.11


Date: 01/13/99 at 20:10:20
From: Sam
Subject: A+B+C+D=7.11 A*B*C*D=7.11

Here is the problem: Four numbers ($2.58 or $.14 - this kind of number) 
when added or multiplied equal $7.11. The numbers you use and the 
answer may only be in money units, with no rounding at all. 

I think you must use two values to get a divisor of 7.11.

Thanks for any help you can give me,
Sam 


Date: 01/14/99 at 11:35:42
From: Doctor Rob
Subject: Re: A+B+C+D=7.11 A*B*C*D=7.11

Thanks for writing to Ask Dr. Math!

This is fun, but a bit complicated.

Express all the amounts of money in cents, instead of dollars. Call the 
amounts of money A, B, C, and D, which are all natural numbers. Then

   A + B + C + D = 711
   A*B*C*D = 711000000
   1 <= A < 711
   1 <= B < 711
   1 <= C < 711
   1 <= D < 711

First observe that the arithmetic mean of A, B, C, and D is

   (A+B+C+D)/4 = 711/4 = 177.75

and their geometric mean is

   (A*B*C*D)^(1/4) = 711000000^(1/4) = 163.29

It is a useful fact that the geometric mean (g.m.) of any set of 
positive real numbers is always no greater than the arithmetic mean 
(a.m.).

Now factor 711000000 into prime power factors

   A*B*C*D = 2^6 * 3^2 * 5^6 * 79

One of them must be divisible by 79. Say it is A that is divisible by 
79. Then A = 79*a, where 1 <= a < 9, and a | 2^6 * 3^2 * 5^6.

The fact that the arithmetic mean of B, C, and D is greater than their
geometric mean gives

(B+C+D)/3 = (711-79*a)/3 < (B*C*D)^(1/3) = [711000000/(79*a)]^(1/3)
493039*a^4 - 13312053*a^3 + 119808477*a^2 - 359425431*a + 243000000 > 0

This equation has solutions

   0.94197... < a < 4.04065...

Because a is an integer, we have 1 <= a <= 4. Then

   B + C + D = 711 - 79*a <= 632 < 3125 = 5^5

so 5^5 cannot divide B, C, or D.

Suppose first that B + C + D is not divisible by 5. Then exactly one of 
them is not divisible by 5, and we label this one B. If 625 = 5^4 
divides either C or D, say C, then 625 <= C <= 711 < 2*625, so C = 625. 
Then B + D = 86 - 79*a >= 2, which implies a = 1, B + D = 7, and 
B*D = 14400. Then the a.m. of B and D is 3.5 and their g.m. is 120, 
which is impossible. Thus neither C nor D can be divisible by 5^4, 
so both C and D must be divisible by 5^3 = 125. Write C = 125*c, 
D = 125*d. Then we have B + 125*(c+d) + 79*a = 711, a*B*c*d = 9. 
Then c + d < (711-79-1)/125, so 2 <= c + d <= 5.  Also a, B, c, and d 
can only be 1, 3, or 9, so one of c and d must be 1, say c, and then d 
is 1 or 3. Now a*B*d = 9, B + 125*d + 79*a = 632. If d = 1, then 
B + 79*a = 461 and a*B = 9, which lead to the quadratic equation 
79*a^2 - 461*a + 9 = 0, which has no rational roots, a contradiction.  
If d = 3, then B + 79*a = 211 and a*B = 3, which lead to the quadratic 
equation 79*a^2 - 211*a + 3 = 0, which also has no rational roots, a 
contradiction. So B + C + D not divisible by 5 is impossible.

Thus B + C + D must be divisible by 5, which implies that a = 4,
A = 4*79 = 316, B + C + D = 395, B*C*D = 2250000.

Now by an argument similar to the above, all three of B, C, and D must 
be divisible by 5, so put B = 5*b, C = 5*c and D = 5*d, which gives

   b + c + d = 79
   b*c*d = 18000 = 2^4 * 3^2 * 5^3

As above, just two of b, c, and d are divisible by 5, and one of them 
by 5^2. Say

   b = 25*x,  c = 5*y,  25*x + 5*y + d = 79,  x*y*d = 2^4 * 3^2 = 144

Now x <= (79-5-1)/25, so 1 <= x <= 2. If x = 2, we get

   5*y + d = 29,  y*d = 72

and the a.m. of 5*y and d is 14.50, and their g.m. is 18.97, which is
impossible. Thus x = 1, and

   5*y + d = 54,  y*d = 144

This leads to the quadratic equation

   5*y^2 - 54*y + 144 = 0
   (5*y - 24)*(y - 6) = 0
   y = 6 or y = 24/5

The last root is not an integer, so it is extraneous, and you get the
unique solution

   y = 6
   x = 1
   d = 24
   c = 30
   b = 25
   a = 4
   A = 316
   B = 125
   C = 150
   D = 120

and the amounts of money were $3.16, $1.50, $1.25, and $1.20.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   

Associated Topics:
High School Number Theory
High School Puzzles

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