Numbers game

Observe that $1000 = 2^{3} \times 5^{3}$ . Therefore, if we group these prime factors into 4 numbers, then we know that exactly 1 of each of the 4 composite numbers $c_{1}, c_{2}, c_{3}, c_{4}$ is divisible by exactly one of each of the 4 numbers from the prime factorization of 1000 because

$\\ \frac{c_{1} \times c_{2} \times c_{3} \times c_{4}}{2^{3} \times 5^{3}} = p_{1} \times p_{2} \times p_{3} \times p_{4}$

We now must find $1000 = a \times b \times c \times d$ such that $c_{1} = a \times p_{1}$ , $c_{2} = b \times p_{2}$ , $c_{3} = c \times p_{3}$ , $c_{4} = d \times p_{4}$

By chosing $a, b, c, d$ so that there sum is as small as possible will yield the smallest composite numbers. However when we choose $a, b, c, d$ to be $8, 5, 5, 5$ the repetition of the digit 5 means that instead of multiplying by the smallest prime number, 2, we have to each time multiply by the next biggest prime number. Therefore we let $a = 10, b = 5, c = 5, d = 4$ .

Therefore,

$c_1 = 10p_{1} = 10 \times 2 = 20 \\ c_2 = 5p_{2} = 5 \times 2 = 10 \\ c_3 = 5p_{3} = 5 \times 3 = 15 \\ c_4 = 4p_{4} = 4 \times 2 = 8$

Therefore, the sum of the composite number is

$\\ \sum _{ i=1 }^{ 4 }{ c_{ i } } = 20 + 10 + 15 + 8 = 53$

So the minimum possible value of $c_{1} + c_{2} + c_{3} + c_{4}$ is 53.