Finding Ratios and First Term of Geometric Sums

cake27

Three integers x, y, and z (where x <= y <= z) form a geometric series with a common ratio r (which is also an integer). Given x + y + z = 741, what is the sum of all possible values of z? (Ans Key: 4870)

I started doing some guess and check with my calculator and got z values of 247, 513, and 637 so far (with respective common ratios of 1, 3, and 7). These values add up to 1397, but I'm wondering if there's a quicker way. Any thoughts?

DenisB

I also get 1397; are you sure AnsKey = 4870?

You do know that y = xr and z=xr^2, right?
So x + xr + xr^2 = 741

13+91+637 = 741
57+171+513 = 741
247+247+247 = 741

Archie

$$\displaystyle x(1+r+r^2) = 741$$

Prime factors of $$\displaystyle 741$$ are $$\displaystyle 3$$, $$\displaystyle 13$$, $$\displaystyle 19$$, so $$\displaystyle x \in \{1, 3, 13, 19, 39, 57, 247, 741\}$$. That's only eight cases to look at.
\displaystyle \begin{aligned} x=741 \implies r^2+r+1 &= 1 \implies r \in \{0,-1\} \implies z \in \{0, 741\} \\ x=247 \implies r^2+r+1 &= 3 \implies r \in \{1,-2\} \implies z \in \{247, 988\} \\ x=57 \implies r^2+r+1 &= 13 \implies r \in \{3,-4\} \implies z \in \{513,912\} \\ x=39 \implies r^2+r+1 &= 19 \qquad \text{has no integer solutions} \\ x=19 \implies r^2+r+1 &= 39 \qquad \text{has no integer solutions} \\ x=13 \implies r^2+r+1 &= 57 \implies r \in \{7,-8\} \implies z \in \{637, 832\} \\ x=3 \implies r^2+r+1 &= 247 \qquad \text{has no integer solutions} \\ x=1 \implies r^2+r+1 &= 741 \qquad \text{has no integer solutions} \end{aligned}​

Thus we have $$\displaystyle z \in \{0,247,513,637,741,832,912,988\}$$ and $$\displaystyle 0+247+513+637+741+832+912+988=4870$$

I'm guessing you guys missed the negative ratios.

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cake27

Thanks! I didn't realize the fact that I only needed to check the factors of 741 (silly me (Itwasntme)).