# Problem of the Day: 3/12/13

In how many different ways can 10 people be seated at a round table? Note that arranging the people the same way at different positions around the table does not count.

Solution to yesterday’s problem:

Since the common factor of the two numbers is 4, both numbers are divisible by 4. We can therefore say that the first number is 4 times a number, or 4m. The second number is 4 times another number, or 4n. The problem tells us that

$4m+4n=52$

Factoring out a 4 on the left side,

$4(m+n)=52$

We can now divide by 4:

$m+n=13$

So the range of possible values for m and n is such that their sum is 13: 1 and 12, 2 and 11. However, logic tells us that since we want the smallest possible difference between the numbers, we should choose numbers close to 13/2; namely, 6 and 7. Using 6 and 7, our original numbers would be 24 and 28. These do have a sum of 52 and a GCF of 4, and their difference is 4.