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### Adding and Multiplying to Get 7.11

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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.

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 inequality 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

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/

```

```
Date: 05/17/2017 at 05:07:43
From: Ines Mueller
Subject: Adding and Multiplying to Get 7.11

Dear Dr. Math,

I have a very similar problem: find four amounts that when added or
multiplied equal 7.35.

I tried to solve my problem by referring to the 1999 conversation above,
where the target was 7.11.

I started with this:

A + B + C + D = 735
(A + B + C + D)/4 = 735/4
= 183.75

A*B*C*D = 735000000
(A*B*C*D)^(1/4) = 164.65

But then factorization problems began ...

I do not understand how Doctor Rob come to the seemingly important
conclusion that "one of them must be divisible by 79." Nor do I understand
the next formula.

Would you assist me?

Thanks in advance and greetings from Switzerland,

Ines
```

```Date: 05/17/2017 at 23:21:36
From: Doctor Peterson
Subject: Re: Adding and Multiplying to Get 7.11

Hi, Ines.

The 79 comes from when Doctor Rob factored 711000000. He found that the
largest prime factor is 79. Since that is a factor of A*B*C*D, it must be
a factor of one of A, B, C, and D.

In your case, you would want to factor your problem's analogous number,
(735000000), and hope that it will provide a similarly notable prime
(it doesn't).

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

Here, he is using the inequality "arithmetic mean > geometric mean" to
write this:

(B + C + D)/3 > (B*C*D)^(1/3)

(711 - 79*a)/3 > [711000000/(79*a)]^(1/3)

If we try solving this inequality, we might do this:

(711-79*a)/3 > [711000000/(79*a)]^(1/3)

(711-79*a) > 3*[711000000/(79*a)]^(1/3)

(711-79*a)^3 > 27*711000000/(79*a)

a(711-79*a)^3 > 27*711000000/79

243000000 - a(711-79*a)^3 < 0

This expands to

493039a^4 - 13312053a^3 + 119808477a^2 - 359425431a + 243000000 < 0

This is Doctor Rob's inequality.

When I ask Wolfram|Alpha to solve this, it says

0.941974 < a < 4.04065

And that is just what he said. (I suspect he, too, used software to help!)

Does this help you understand the work, so you can try using a similar
method for your problem? (I can't promise that everything will work out
the same with different numbers.)

- Doctor Peterson, The Math Forum at NCTM
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory
High School Puzzles

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