The Lemon Fool

348 numbers are written round a circle such that the sum of any 20 consecutive numbers is 95. The numbers 3, 4 and 7 are in positions 21, 123 and 288 respectively. What is in position 310?

Cinelli

Spoiler:

5

Because every sum of 20 consecutive numbers is 95 the number pattern must repeat every 20 positions.

Because 348 is divisible by 4 but not by 5 the number pattern must repeat every 4 positions.

Thus A[n] = A[mod(n,4)]
We have only four distinct numbers.
mod(21,4) = 1 ==> A[1] = 3
mod(123,4) = 3 ==> A[3] = 4
mod(288,4) = 0 ==> a[0] = 7

Also 5 * (A[0]+A[1]+A[2]+A[3]) = 95
so A[2] = 5

mod(310,4) = 2 ==> A[310] = A[2] = 5

cinelli wrote:348 numbers are written round a circle such that the sum of any 20 consecutive numbers is 95. The numbers 3, 4 and 7 are in positions 21, 123 and 288 respectively. What is in position 310?

Spoiler (posted before reading 9873210's spoiler)...

Consider 21 consecutive numbers a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u. We're told that a+b+c+d+e+f+g+h+i+j+k+l+m+n+o+p+q+r+s+t = 95 = b+c+d+e+f+g+h+i+j+k+l+m+n+o+p+q+r+s+t+u, from which it follows that a = u, i.e. that numbers 20 places apart are equal to each other.

So if we choose to call a particular number the 1st number around the circle, it is equal to the 21st number, which is equal to the 41st number, which is equal to the 61st number, ..., which is equal to the 361st number, which is the 13th number because we've wrapped around the 348-place circle. Continuing in the same vein, the 13th number is equal to the 33rd number, the 53rd number, the 73rd number, ... and the 353rd number, which is the 5th number because we've wrapped around again. So starting with any number as the 1st number, we know that it's the same as the 5th number, i.e. that numbers 4 places apart are equal to each other.

So the 1st, 5th, 9th, 13th, ... and 345th numbers are all equal to each other, and since one of them is the 21st number, they're all equal to 3. Similarly, the 2nd, 6th, 10th, 14th, ... and 346th numbers are all equal to the 310th number, i.e. the number N that we're after. And the 3rd, 7th, 11th, 15th, ... and 347th numbers are all equal to 4, the 123rd number, and the 4th, 8th, 12th, 16th, ... and 348th numbers are all equal to 7, the 288th number.

So any 20 consecutive numbers are 5 copies of 3, 5 copies of N, 5 copies of 4 and 5 copies of 7, and so 15+5N+20+35 = 95, from which it follows that N = 5.

Edit: Having now read 9873210's spoiler, it's essentially the same as mine (which doesn't surprise me) but expressed more 'mathematically' than I chose to!

Gengulphus

Well solved, both. It's nice when two slightly different approaches give the same answer.

Cinelli

Gengulphus wrote:Edit: Having now read 9873210's spoiler, it's essentially the same as mine (which doesn't surprise me) but expressed more 'mathematically' than I chose to!

Gengulphus

I don't think my answer was more mathy. I did a lot of handwaving. Partly because of the inelegance of expressing "sum from i=0 to 19 ... " in a medium with poor access to Greek letters, sub scripts, and super scripts; but mostly because I'm lazy. If I was pushed to clarify I'd replace the handwaving much as you did.

If you want a really different approach set up the 351 simultaneous equations in 348 variables and left invert the matrix. That was my first thought, but my second thought was that if it took that much work the problem could remain unsolved.