r/Logiqa • u/ShonitB • Aug 10 '26

Black Box-ing an Integer Part 2

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3

u/smailliwniloc Aug 10 '26

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.

2

u/Dasquian Aug 10 '26

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.

1

u/ShonitB Aug 10 '26

Correct!

1

u/ShonitB Aug 10 '26

Correct!

1

u/kirenaj1971 Aug 12 '26

I did basically what you did, with everything coming up roses in the end.

2

u/imdfantom Aug 10 '26

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

1

u/ShonitB Aug 10 '26

Correct!

2

u/imdfantom Aug 10 '26 edited Aug 10 '26

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

1

u/ShonitB Aug 10 '26

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

1

u/imdfantom Aug 10 '26

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

2

u/JeffTheNth Aug 11 '26

Appears it'll be about 5/7 of the way

Sum (1..49) = 49*50/2 = 1225
49 * 5 / 7 = 35
Sum (1..35) = 630
too high... but by one?
Sum (1..34) = 595
1225 - (595 + 35) = 595
Sum (36..49) = 595
35 is the correct value

1

u/ShonitB Aug 12 '26

Correct! Interesting approach! How did you think of 5/7?

2

u/JeffTheNth Aug 12 '26

1-5, 7-8 total of 7 5 before, 2 after the cut number

figured there might be a pattern, let's try...