Number Theory?

Question

Anna says, "We three have $100 altogether." Betty says, "Yes, and if you had six times as much and I had one-third as much, we three would still have $100." Carl says, It's not fair. I have less than $30." Who has what? (Dudley)

Answer 1

Anna = 10
Betty = 75
Carl = 15

Anna = x, Betty = y, Carl = z

x + y + z = 100
6x + y/3 + z = 100
0 < z <30

From the first 2 equations, we get
x + y = 6x + y/3
=> x = 2y/15

Substitute in either of the first two equations to get:
17y/15 + z = 100
=> z = 100 - 17y/15

Now, as per the third equation,
0 < z < 30
=> 0 < 100 - 17y/15 < 30
=> 0 < 1500 - 17y < 450
=> 1050 < 17y < 1500
=> 61.76 < y < 88.24

Now, y is a multiple of 15 since 17y/15 + z = 100, else the numbers will not be natural numbers.

Hence, the only number that satisfies the inequality which is a multiple of 15 is 75.

Hence, the numbers 10, 75 and 15.

Answer 2

The problem gives you three pieces of information:
a = Anna, b = Betty, c = Carl
(1) a + b + c = 100
(2) 6a + b/3 + c = 100
(3) c<30

However, only Anna's and Betty's money changes between scenario (1) and scenario (2). Carl's money stays constant. This means we can subtract Carl's money from both sides of the equation, and because a constant (100) minus a constant (Carl's money) equals a constant, we can rename the difference of the two constants "K."

a + b = 100 - c
6a + b/3 = 100 - c

a + b = k
6a + b/3 = k

Now, just solve using systems of equations. Solving for a:
a + b = k
[6a + b/3 = k ] x 3

a + b = k
18a + b = 3k

-a - b = -k
18a + b = 3k

17a = 2k
a = 2k/17

Solving for b:
a + b = k
b = 17k/17 - 2k/17
b = 15k/17

If this problem didn't mention deal with money, there would be an infinite number of solutions to this problem. But since we are dealing with money, there is a restriction on the possible answers from which we can choose: repeating and non-repeating decimals can't be the correct answer. That means that a, b, and c must be an answer that ends with a terminal decimal in the hundredths place.

We know that a = 2k/17. We know that b = 15k/17. The denominator is a prime number. This means that the only way for "a" and "b" can be a terminal decimal is if the numerator (2k and 15k) is divisible by 17. So let's start looking for multiples of 17. Let's start with the first six multiples of 17.

(1) k = 1 x 17 = 17
(2) k = 2 x 17 = 34
(3) k = 3 x 17 = 51
(4) k = 4 x 17 = 68
(5) k = 5 x 17 = 85
(6) k = 6 x 17 = 102

Remember that k = 100 - c. And that c<30. Let's see what values we get for c for each of the six values of k we came up with above. If we rearrange k = 100 - c by solving for c, we get c = 100 - k.

c = 100 - k
(1) c = 100 - 17 = 83
(2) c = 100 - 34 = 66
(3) c = 100 - 51 = 49
(4) c = 100 - 68 = 32
(5) c = 100 - 85 = 15
(6) c = 100 - 102 = -2

Remember that we are dealing with money. You can't have -2 dollars, so choice (6) can be eliminated. Also remember that c<30, which eliminates answer choices (1) through (4). Only answer choice (5) satisfies the condition that c<30. According to answer choice (5), c = $15. Carl has $15. According to answer choice (5), k = 85. Let's plug that back into the equation to solve for a and b.

c = $15
k = 85
a = 2k/17 = 2(85)/17 = $10
b = 15k/17 = 15(85)/17 = $75

So our preliminary guess is that Anna has $10, Betty has $75, and Carl has $15. Finally, let's plug this into the original equations to double-check our answers.

Anna says, "We three have $100 altogether."
a + b + c = $100
$10 + $75 + $15 = 100
$100 = $100

Good. The first condition was satisfied. Let's move on to the second condition: "Betty says, 'Yes, and if you had six times as much and I had one-third as much, we three would still have $100.'"
6a + b/3 + c = $100
6($10) + $75/3 + $15 = $100
$60 + $25 + $15 = $100
$100 = $100

So far so good. Finally, let's see if the third and final condition is met: "Carl says, It's not fair. I have less than $30.'"
c<$30
c = $15

Check, check, and check. This means our preliminary guess is a viable answer.

Anna has $10.
Betty has $75.
Carl has $15.

Answer 3

a + b + c = 100
6a + b / 3 + c = 100

Subtracting the 1st equation from the 2nd equation gives :
5a - 2b / 3 = 0
or, 15a = 2b
or, b = 15a / 2
This shows that 'a' must be even for 'b' to be an integer

Substituting 15a / 2 for b, in a + b + c = 100 gives :
a + 15a / 2 + c = 100
or, c = 100 - 17a / 2

But c < 30,
therefore, 100 - 17a / 2 < 30
or, a > 140 / 17 > 8.235...
But as we've seen, 'a' must be even. Therefore, a >= 10
If a = 10, then b = 15*10 / 2 = 75.
Therefore, c = 100 - 10 - 75 = 15

The next value for 'a' is a = 12,
which means b = 90, but a + b = 12 + 90 > 100,

so there is only one solution : a = $10, b = $75 and c = $15.

Answer 4

An example of linear programming.

Note Anna amount must be less than 16, Betty amount must be a multiple of 3 and Carl is less than 30, by trial and error I found two solutions

Anna 12 Betty 84 Carl 4 or
Anna 13 Betty 66 Carl 21

Answer 5

I assume the amount that each person has is in increments of whole dollars.

We have

a + b + c = 100
6a + (1/3)b + c = 100
c < 30

Subtracting the first equation from the second:

5a - (2/3)b = 0
5a = (2/3)b
15a = 2b

Possible solutions in whole dollars

a...b...c
2 15 87
4 30 66
6 45 49
8 60 32
10 75 15

Only the last solution has c < $30. So it is the correct one.

Anna has $10
Betty has $75
Carl has $15

Answer 6

Any split such that:

Carl < 30
(6 * Anna) + (Betty / 3) = Anna + Betty

Here is one (of many) possible solutions:

(Carl, Anna, Betty) = (15, 10, 75)

Answer 7

a + b + c = 100
6a + 1/3 b + c = 100
c < 30

this should be solved for the answer on your question.
3a + 3b + 3c = 300
18a + b + 3c = 300

15a -2b = 0
and c < 30

a = 2 b = 15 c = ... too big
a= 6 b =10 c = .. still too big
a=10 b = 75 c = 15 Yesss
a=8 b = 60 c .. too big
a = 12 b = 90 c would be negative

so only : a=10 b = 75 c = 15 Yesss

Answer 8

Let's see if I can translate the problem:

a+b+c = 100
6a+b/3+c = 100
c<30

I picked c=20, and solved (a,b) = (9.41,70.59)

For c=25, (a,b) = (8.82,66.18) another solution

So, there are lots of solutions. Since you said 'number theory', then I assume that you want whole numbers.

That solution is (a,b,c) = (15,75,10)

I remember doing these problems in my number theory class. I think that they're know as 'Chinese Remainder Therorem' problems. I don't want to go look for my notes but I multiplied the second equation by 3 and then prime factored all coefficients:

The only prime factors needed are: 2, 3, 5

Note that the solution is made has numbers with these prime factors.

Answer 9

$100 cannot be divided evenly among 3 people.