Geometric progression

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Doppel

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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.
 
You say only three 5's but there could be eleven 7s. Just say that no power (just not for 5) is at least 6 OR all powers are less than 6.
 
Doppel said:
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? 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.
Click to expand...
I find this solution completely correct.
The confusion may by in the factorization \(7875=3^2\cdot 5^3\cdot 7\). SEE HERE
Using \(a\) for the first term & \(r\) for the common ratio them a four term gp is \(a,ar,ar^2,ar^3\) the product of terms is \(a^4r^6\).
Can that product divide \(7875~?\)
 
Doppel said:
What if comman ratio is rational number?
Click to expand...
That's an interesting thought. It is possible for all terms of a (finite) geometric sequence to be integers though the common ratio is not; for example, 27, 18, 12, 8, with common ratio 2/3. Of course, in this case the product is [MATH]2^6\cdot 3^6[/MATH], just as if it had been 1, 6, 36, 216.

Can you see how to extend your initial thinking to cover this case?
 
[QUOTE = "Доктор Петерсон, должность: 499655, участник: 62318"]
Это интересная мысль. Все члены (конечной) геометрической последовательности могут быть целыми числами, хотя общее отношение не является; например, 27, 18, 12, 8, с общим соотношением 2/3. Конечно, в этом случае произведение будет [MATH] 2 ^ 6 \ cdot 3 ^ 6 [/ MATH], как если бы это было 1, 6, 36, 216. Можете ли вы увидеть, как расширить свое первоначальное мышление, чтобы охватить этот случай? [/ QUOTE] [ATTACH type="full"]19875[/ATTACH][/MATH]
 
I'm not sure what your last three lines mean; some words would help. What do you mean by min(n^3)?

What I would focus on is that n^3 | a, and what that implies about the product.
 
[QUOTE = "Доктор Петерсон, должность: 499809, участник: 62318"]
Я не уверен, что означают твои последние три строки; некоторые слова помогут. Что вы подразумеваете под мин (п ^ 3)?

На чем я бы остановился, так это на том, что n ^ 3 | и что это означает о продукте.
[/ QUOTE]
Min(n^3) is a >= n^3
 
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