1. Number line problem

On a number line, tick marks are spaced so that the distance from a to b is twice the distance from b to c and the distance from b to c is twice the distance from c to d. If a, b, c, and d are positive integers, what is the smallest possible value of d?

How can we show that it is 8?

2. Re: Number line problem

Originally Posted by mjoshua
On a number line, tick marks are spaced so that the distance from a to b is twice the distance from b to c and the distance from b to c is twice the distance from c to d. If a, b, c, and d are positive integers, what is the smallest possible value of d?

How can we show that it is 8?
1. The smallest possible tick is at 1. Thus a = 1.

2. Let x denote the distance between a and b: b = a + x

3. Then you get:

$\displaystyle{d = a + x + \frac12 x + \frac12 \left(\frac12 x \right) = 1 + \frac74 x = \frac{7x+4}4}$

4. Obviously the quotient is a positive integer if x = 4, therefore d = 8.

3. Thought twice

Originally Posted by mjoshua
On a number line, tick marks are spaced so that the distance from a to b is twice the distance from b to c and the distance from b to c is twice the distance from c to d. If a, b, c, and d are positive integers, what is the smallest possible value of d?

How can we show that it is 8?
Could it be that you didn't mention that a < b < c < d ?

Since the distance is a scalar (and not a vector) the ticks could be arranged:
Code:
0--1--2--3--4--5\\
---a--d--c-----b
In this case the smallest value of d = 2.

Maybe there other arrangement possible (?)

4. Re: Number line problem

Hello, mjoshua!

Did you make a sketch?

On a number line, tick marks are spaced so that:
. . the distance from a to b is twice the distance from b to c
. . and the distance from b to c is twice the distance from c to d.
If a, b, c, and d are positive integers, what is the smallest possible value of d?

How can we show that it is 8?

The number line looks like this:

. . $\begin{array}{cccccccc} & 4x && 2x && x \\ \bullet & ----- & \bullet & --- & \bullet & -- & \bullet \\ a && b && c && d \end{array}$

Since we want the smallest value of $d$, let $x \,=\,\overline{cd} \,=\,1$
. . and let $a = 1$, the smallest positive integer.

Then we have:

. . $\begin{array}{cccccccc} & 4 && 2 && 1 \\ \bullet & ----- & \bullet & --- & \bullet & -- & \bullet \\ 1 && 5 && 7 && 8 \end{array}$

Therefore: . $d\,=\,8$

Earboth,
And I'm impressed that you found a solution . . .

5. Re: Number line problem

Originally Posted by Soroban
[size=3]...

Earboth,