The fractions 3/5, 7/11, 11/17, 15/23, ... follow a fixed pattern. One later term is A/305 and another is 1999/B. Find the greatest common divisor of A and B.

Step-by-step solution

1. 1

   Treat the numerators and denominators as two separate sequences. Numerators: $3,7,11,15,\ldots$ increase by $4$ each time. Denominators: $5,11,17,23,\ldots$ increase by $6$. Both are arithmetic, so each has a closed form:

   $\text{numerator}_n=3+4(n-1)=4n-1$

   $\text{denominator}_n=5+6(n-1)=6n-1$

   Spot check at $n=4$: $4(4)-1=15$ and $6(4)-1=23$ $\checkmark$.

2. 2

   Locate the term whose denominator is 305.

   $6n-1=305\ \Longrightarrow\ 6n=306\ \Longrightarrow\ n=51$

   That $n$ came out a whole number confirms $305$ really is in the sequence. Its numerator is

   $A=4(51)-1=203$

3. 3

   Locate the term whose numerator is 1999.

   $4n-1=1999\ \Longrightarrow\ 4n=2000\ \Longrightarrow\ n=500$

   $B=6(500)-1=2999$

4. 4

   Factor both numbers before assuming a common divisor.

   $203=7\times 29$

   For $B=2999$, trial division by every prime up to $\sqrt{2999}\approx 54.8$ — $2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53$ — leaves a remainder every time, so $2999$ is prime.

5. 5

   Compute the GCD. Since $2999$ is prime, the only candidates for a shared factor are $1$ and $2999$ itself, and $2999\nmid 203$. Confirming with the Euclidean algorithm:

   $2999=14\times 203+157,\quad 203=1\times 157+46,\quad 157=3\times 46+19$

   $46=2\times 19+8,\quad 19=2\times 8+3,\quad 8=2\times 3+2,\quad 3=1\times 2+1,\quad 2=2\times 1$

   $\gcd(203,2999)=1$

   The two numbers are coprime. (It is tempting to write $2999=29\times 103$ and conclude the answer is $29$, but $29\times 103=2987$, not $2999$ — a factorisation worth checking by multiplying back.)