Math Puzzle - Circa1953?

Question

First number is twice as large as the third.
First number and the third number are equal to the second number.
Half of the second number and the whole of the third equals the first number.

Answer 1

Let the numbers be A, B & C.


First number is twice as large as the third.
A = 2C

First number and the third number are equal to the second number.
A + C = B

Half of the second number and the whole of the third equals the first number.
B/2 + C = A

A = 2C ..... (1)

A + C = B ..... (2)

B/2 + C = A ..... (3)

Substitute (1) into (2),

[2C] + C = B

3C = B ..... (4)

Substitute (1) into (3)

B/2 + C = [2C]

B/2 = 2C - C

B/2 = C

B = 2C ..... (5)

Substitute (5) into (4)

3C = [2C]

3C - 2C = 0

C = 0

if C=0, then A=0 from (1)

if C=0 and A=0, then B=0 from (2)


Hence,

A = 0
B = 0
C = 0

Answer 2

So, our three rules are

n1 = 2*n3
n1 + n3 = n2
1/2*n2 + n3 = n1

Restating these:
n1 = 2*n3
n1 = n2 - n3
n1 = 1/2*n2 + n3

The only way for these to be true is if n1 = n2 = n3 = 0.

Answer 3

Who is the MEAN teacher that gave you this problem for homework?