MHB Finding Minimum Value of $n$ for Given Sum and Product

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To find the minimum value of \( n \) such that the sum and product of \( n \) integers equal 2006, the integers must be positive and their sum must equal their product. The discussion highlights the relationship between the sum and product of integers, emphasizing that smaller integers are preferable to minimize \( n \). It is noted that the prime factorization of 2006 is \( 2 \times 7 \times 143 \), which can be further broken down into \( 2 \times 7 \times 11 \times 13 \). The minimum \( n \) is determined through trial and error with combinations of these factors, leading to the conclusion that \( n \) must be at least 4. The final result indicates that the minimum value of \( n \) is 4.
Albert1
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$n\in N,\,\,and \,\, a_1,a_2,a_3,-------,a_n\in Z$
$if \,\, a_+a_2+a_3+-----+a_n=a_1\times a_2\times a_3\times------\times a_n=2006$
$find \,\, min(n)$
 
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My attempt:

The prime factorization of $2006$ is $2\cdot17\cdot59$. This leaves very few possibilities to express $2006$ as a product of integers:(i). $2\cdot17\cdot59$. Sum of factors: $78$.

(ii). $34\cdot59$. Sum of factors: $93$.

(iii). $17\cdot118$. Sum of factors: $135$.

(iv). $2\cdot1003$. Sum of factors: $1005$.

(v). $1 \cdot 2006$. Sum of factors: $2007$.

In order to obtain the expression:
$a_1\cdot a_2…\cdot a_n = a_1+a_2+…+a_n = 2006$, we only have the factors mentioned in (i)-(v) and the neutral factor $\pm 1$ to fill out with. So, the task is to use a factorization, which involves the fewest number of $1$´es.

Keeping this in mind, the lowest number of factors/terms is obviously obtained, when the sum of factors is largest (case (v).). Consequently, we are aiming for the identity:

$(-1) + 1 + (-1) + 1 + 2006 = (-1) \cdot 1 \cdot (-1) \cdot 1 \cdot 2006$

Hence, the minimum number of factors/terms is $n_{min} = 5$.

P.S.: Case (iv). would imply $n = 1003$.
 
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