My, What Large Numbers You Are

The number should be odd since there are nunbers in the sequence that are not even. Evaluate $a_{1} = 2\cdot 4\cdot 6\cdot 8\cdot 10 = 2^{8} \cdot 15$ . Im going to prove that in any term of the sequence there will appear the factors $3$ and $5$ . The final digits can be one of these: in some order: $1, 3, 5, 7, 9$ or $2, 4, 6. 8, 0$ . Obviously the ones enting with $5$ and $0$ are divisible by $5$ . But what about three? We can subtitute some $n$ of the forms $3x$ $3x + 1$ or $3x + 2$ . For the first case the second term would give $3x + 3$ . The second case, the third term gives $3x + 6$ . For the third case, the fifth term gives $3x + 9$ . Therefore there will always appear the factors $3$ and $5$ and thus all the terms will be divisible by the answer: $\boxed{15}$ .

if the nth term is expressed as = (n+1)(n+3)(n+5)(n+7)(n+9) , then the first term can be expressed as = (1+1)(1+3)(1+5)(1+7)(1+9) which has prime factorization 2,2,2,2,3,2,2,2,2 and 5 and then the second term of this sequence can be written as = (2+1)(2+3)(2+5)(2+7)(2+9) which has prime factorization 3,5,7,9,11 then we find out the common factors between the prime factorization of first and second term of a sequence which would be the largest common integer that is 3 * 5 = 15