<s>[INST] The sum of ten distinct natural numbers is $62$. Show that their product is a multiple of $1440$. 
The sum of ten distinct natural numbers is $62$. Show that their product is a multiple of $1440$.

I know how to prove that it is a multiple of $60$
Lets write: $a_1+a_2+...+a_{10}=62$. Assuming $a_1<a_2<\cdots <a_{10}$ we search the maximal value of $a_{10}$. This is the difference of $62$ and the minimal value the sum  $a_1+a_2+\cdots+a_9$ can take. As this value is given by $1+2+\cdots+9=45$, it follows that $a_{10}=62-45=17$. That means that for any $a_i>17$, there are no $9$ natural numbers to satisfy our initial sum.
So a solution would be $1,2,3,4,5,6,7,8,9,17$.
If $a_{10}=16$, the unique solution is $a_9=10$, the smaller solutions being unchanged as to preserve the distinctness.
If $a_{10}=15$, then $a_9=11$ or $a_8=10$.
If $a_{10}=14$, then $a_9=12$ or $a_8=11$ or $a_7=10$.
If $a_{10}=13$, then $a_8=12$ or $a_7=11$ or $a_6=10$.
If $a_{10}=12$, $a_9=13$ or $a_6=11$ or $a_5=10$. And from here on, the situation repeats.
We observe that the numbers $3,4,5$ or $6,10$ or $5,12$ repeat and as their products are $60$, then $60$ divides the product of $a_1,\cdots,a_{10}$. I don't know how to expand for $1440$. Any help, please?
 [/INST] Experimental approach:
Sum of first 11 numbers is:
$S=\frac {11}2(11+1)=66$
$66-62=4$
So we delete 4 and our numbers are:
$1, 2, 3, 5, 6, 7, 8, 9, 10, 11$
which their product is a multiple of 1440.
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