# Number line problem

• Jul 7th 2012, 10:41 PM
mjoshua
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?
• Jul 7th 2012, 11:03 PM
earboth
Re: Number line problem
Quote:

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:

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

4. Obviously the quotient is a positive integer if x = 4, therefore d = 8.
• Jul 8th 2012, 03:47 AM
earboth
Thought twice
Quote:

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 (?)
• Jul 8th 2012, 07:31 AM
Soroban
Re: Number line problem
Hello, mjoshua!

Did you make a sketch?

Quote:

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:

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

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

Then we have:

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

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

Earboth,
And I'm impressed that you found a solution . . .
• Jul 8th 2012, 08:14 AM
earboth
Re: Number line problem
Quote:

Originally Posted by Soroban
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Earboth,