Question 536865
If only that poor commander had 15 more soldiers, he would have choices. He could make 70 equal rows, or 65 equal rows, or 60 equal rows, without any soldiers left over. Do you see why? In each case, the soldiers left over plus 15 more soldiers would equal the number of rows, so you could add one soldier to each row and have no leftovers.
So, if the number of soldiers was {{{n}}},
{{{n+15}}} is a multiple of 60, 65, and 70.
{{{60=2^2*3*5}}} {{{65=5*13}}} {{{70=2-5-7}}}
The smallest common multiple (least common multiple, or LCM) is
{{{2^2*3*5*7*13=5460}}}
That's the only choice, because the next common multiple would be twice that, and it would mean more than 10000 soldiers. So
{{{n+15=5460}}}---> {{{n=5445}}} which is a multiple of 5 and a multiple of 11, and therefore a multiple of 55.
You know why I see at first sight that it's a multiple of 5. To see if it's a multiple of 11, add up two groups of every other digit (ones plus hundreds and tens plus thousands). If the difference between the sums is zero or multiple of 11, the number is a multiple of 11. And if a number is a multiple of two different prime factors, it is a multiple of their product.
So the commander has 5445 soldiers and can line them up in 55 equal rows.