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?
 A: Consider potential minimal counter-examples which might not be multiples of the prime factorisation of $1440=2^53^25$ and show that the sums cannot be $62$:
You must have a multiple of $5$ in the product since $1+2+3+4+6+7+8+9+11+12=63$ is too big and any smaller sum must have a multiple of $5$.
You must have a multiple of $9$ in the product since $1+2+3+4+5+7+8+10+11+13=64$ is too big and any smaller sum must have a second multiple of $3$.
You also want to show you must have a multiple of $32=2^5$ in the product, but that is a little more complicated.  There must be an even number of odd numbers to get an even sum, and since there are $10$ numbers you must have an even number of even numbers.

*

*If you have six or more even numbers, then you will have at least $2^6=64$ in the product

*If you have four even numbers then at least one of them must be a multiple of $4$ so giving $32$ in the product, since $1+2+3+5+6+7+9+10+11+14=68$ is too big

*If you have two or fewer even numbers then the sum will be too big, since $1+2+3+4+5+7+9+11+13+15=70$ is too big

So you must have $5 \times 9 \times 32=1440$ in the product.
You have given an example of $17 \times 9!= 6168960=1440 \times 4284$ and this is in fact the smallest example.
A: 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.
