PRMO 2017 question 16

16. Five distinct 2-digit numbers are in a geometric progression. Find the middle term.

Any integer‐valued 5‐term G.P. of two‐digit numbers forces the common ratio to be either $\tfrac32$ or $\tfrac23$. Indeed, one finds:

• Ratio $\tfrac32$. Starting with $a$, the terms are

$$a,\quad \tfrac32 a,\quad \tfrac94 a,\quad \tfrac{27}{8}a,\quad \tfrac{81}{16}a.$$

All must be integers and two‐digit. In particular, $\tfrac{81}{16}a$ must be an integer, so $a$ must be a multiple of $16$. The only two‐digit multiple of $16$ that keeps all terms under $100$ is $a=16$. That yields

$$16,\;24,\;36,\;54,\;81,$$

whose middle term is $36.$

• Ratio $\tfrac23$. Now the terms are

$$a,\quad \tfrac23 a,\quad \tfrac49 a,\quad \tfrac{8}{27}a,\quad \tfrac{16}{81}a,$$

and similarly $a$ must be a multiple of $81$. Taking $a=81$ stays two‐digit and gives

$$81,\;54,\;36,\;24,\;16,$$

again with middle term $36.$

No other rational ratios can produce five distinct two‐digit integers in G.P. In both valid cases above, the “center” of the progression is $36$. Hence the (unique) middle term is 36.

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In more detail:

A common way to see that the ratio must be $\tfrac32$ or $\tfrac23$ is via a bounding plus integrality argument:

1. The ratio must be rational in lowest terms.
   Suppose our 5‐term G.P. is

   $$a,\quad a r,\quad a r^2,\quad a r^3,\quad a r^4$$

   with each term a two‐digit integer (from 10 to 99). If $r$ were irrational, then even the first two terms $a$ and $ar$ could not both be integers. Hence $r$ must be rational: write

   $$r = \frac{p}{q}$$

   in lowest terms, with $p,q>0$.

2. All terms must be integers.
   From

   $$a,\quad a\frac{p}{q},\quad a\biggl(\frac{p}{q}\biggr)^2,\quad a\biggl(\frac{p}{q}\biggr)^3,\quad a\biggl(\frac{p}{q}\biggr)^4,$$

   we see that for each power of $\tfrac{p}{q}$ to yield an integer when multiplied by $a$, we must have

   $$a \quad \text{is divisible by}\quad q^4.$$

   In symbols, $q^4 \mid a$.

3. Size constraints force small $p$ and $q$.
   Each term must lie in $[10,99]$. In particular,

   $$10 \;\le\; a \;\le\; 99 \quad\text{and}\quad 10 \;\le\; a\bigl(\tfrac{p}{q}\bigr)^4 \;\le\; 99.$$

   Since $a$ is a multiple of $q^4$, you quickly find that only very small values of $q$ are possible (because $q^4$ grows fast and must still divide some integer $a$ in the range 10–99). A similar argument applies to $p$ by looking at the higher‐index terms.

4. Systematic check leaves only $\frac32$ or $\tfrac23.$
   Listing out small rational candidates $\tfrac{p}{q}$ (in reduced form) and testing which ones can fit five distinct two‐digit integers shows that the only possibilities are

   $$r = \frac32 \quad\text{or}\quad r = \frac23.$$

   In both cases, one finds exactly the sequences

   $$16,\;24,\;36,\;54,\;81 \quad\text{and}\quad 81,\;54,\;36,\;24,\;16,$$

   with ratio $\tfrac32$ and $\tfrac23$, respectively. No other ratio can “thread the needle” of being (i) rational in lowest terms, (ii) giving integer‐valued terms, and (iii) keeping all five terms in the two‐digit range.

Hence the forced values of the common ratio are $\tfrac32$ or $\tfrac23$.

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