Te finite geometric progression consist of various natural numbers. The product of the members of this progression is a divisor of the number 7875.
a) Can this progression consist of 3 members?
b) Can this progression consist of 4 members?
c) Can this progression consist of 5 members?
Supplement my answer b,c. Sorry for my english
My solution
a) Prime factorization 7875 = 5^3 * 3^2 * 7 * 1^(n). Answer yes. For example 1, 5, 25.
b,c) If first member of progression is b1, then b4 = b1*q^3. Product of members = b1^4 * q^6. But it is impossible because only three 5. 3< 6.
a) Can this progression consist of 3 members?
b) Can this progression consist of 4 members?
c) Can this progression consist of 5 members?
Supplement my answer b,c. Sorry for my english
My solution
a) Prime factorization 7875 = 5^3 * 3^2 * 7 * 1^(n). Answer yes. For example 1, 5, 25.
b,c) If first member of progression is b1, then b4 = b1*q^3. Product of members = b1^4 * q^6. But it is impossible because only three 5. 3< 6.