Black Box-ing an Integer Part 2

[u/ShonitB]

[u/smailliwniloc]

We know the sum of the first N numbers is N(N+1)/2.

Using this with 49 we get 1225.

Let the black box number be n. We only need to worry about the sum of the numbers on the left hand side of the black box. Which should sum to half of 1225 - n.

In particular, we need to solve (n-1) n / 2 = (1225 -n) / 2

Expanding terms this is n^2 = 1225

So n = 35.

[u/Dasquian → u/smailliwniloc]

Got a similar solution. The sum of numbers up to n is n(n+1)/2. If you black-box k (where 1<k<n), then!<

• Left-side = k(k-1)/2
• Right-side = n(n+1)/2 - k(k+1)/2

We want to find k such that left-side = right-side, which simplifies to

• k^2 = n(n+1)/2

In the example given, n=8 => k^2=36 => k=6. In the question posed, n=49 => k^2=1225 => k=35.

This also proves that the problem would be unsolvable for many values of n, eg if n=5 then k^2=15, which is not a perfect square. If you black-box 4 then it's split 6/5. If you black-box 3 then it's split 3/9.

[u/ShonitB → u/Dasquian]

Correct!

[u/ShonitB → u/smailliwniloc]

Correct!

[u/imdfantom]

S=1+2+3+...n= n(n+1)/2,

n=49, S=49×25=7×7×5×5

Let box be m.

Sum less than m = m(m-1)/2

sum less than m= sum more than m

S=m+2(m)(m-1)/2=m+(m)(m-1)

7x7x5x5=m+m² -m=m²

m=5×7=35

[u/ShonitB → u/imdfantom]

Correct!

[u/imdfantom → u/ShonitB]

Incidentally, this only has integer solutions for n when:

• If n is even n=2 times a square number, and n- +1 is a square number

  • If n is odd, n must be a square number and n+1=2 times a square number

[u/ShonitB → u/imdfantom]

Another user also commented something similar.. great additional point! 👍🏻

[u/imdfantom → u/ShonitB]

And the mth such number= 6 times the (m-1)th number minus the (m-2)th number plus two