Thread by @pgeerkens

@GWOMaths Spoiler - hint and solution follows

@GWOMaths Define Sₖ(c) as the sum of k consecutive integers with offset c thus:
Sₖ(c) = (c+1) + (c+2) + ... + (c+k-1) + (c+k)
= k⋅c + (1 + 2 + ... + (k-1) + k )
= k⋅c + ½k(k+1).

@GWOMaths We desire to find which n ∈ Z⁺ with 50 ≤ n ≤100 can be expressed as Sₖ(c) for some k ≥ 3 and c ≥ 0.

Case 1 - k is odd:
Sₖ(c)
= k ⋅ (c + ½(k+1))
= k ⋅ ½(2c + k + 1)
where ½(2c + k + 1) is integral.

Case 2 - k is even:
Sₖ(c)
= ½k ⋅ (2c + k+1)
where ½k is integral.

@GWOMaths In both cases the value of Sₖ(c) always has an odd factor greater than 1, either k itself in Case 1 or (2c + k+1) in Case 2.

Therefore all powers of 2 are inexpressible as any such Sₖ(c).

Likewise all odd primes are inexpressible as c would be negative.

@GWOMaths There are 11 primes and powers of 2 between 50 and 100 inclusive; namely:
53, 59, 61, 64, 67, 71, 73, 79, 83, 89, 97.

It remains to show that all remaining values of n in the range ARE expressible as such an Sₖ(c).

@GWOMaths For any such n ∈ Z⁺ with 50 ≤ n ≤100:
let p be its smallest odd prime factor (and clearly ≥ 3).

Case: n/p ≥ ½(p+1)

Then in order
2⋅(n/p) ≥ p+1;

2⋅(n/p) - (p+1) ≥ 0;

and since
2⋅(n/p) - (p+1)
is even and positive

then
n = Sₚ( ½(2⋅(n/p) - (p+1)) )
satisfies.

@GWOMaths Case: n/p < ½(p+1)

Note:
If n has any other prime factor, whether p with multiplicity greater than 1 or a q > p, then

n/p ≥ p ≥ ½(p+1)
and we are in the case above.

So this case only occurs when
n = p⋅2ᵐ
for some m > 0.

Then
n/p = 2ᵐ < ½(p+1)

and
2ᵐ⁺¹ < p+1 ≤ p-1.

@GWOMaths So
p - 1 - 2ᵐ⁺¹ ≥ 0

and, since p is odd,
½(p - 1 - 2ᵐ⁺¹)

is integral and positive or zero.

Setting k = 2ᵐ⁺¹ = 2⋅(n/p)
Sₖ(p - 1 - k)
= ½k ⋅ ((p-1-k) + k+1)
= 2ᵐ ⋅ (p)
= n.

Q.E.D.

Note that all positive integers < 6 are either 1; a prime; or a power of 2.

@GWOMaths Therefore all numbers n with 50 ≤ n ≤ 100 are expressible as a sum of at least 3 consecutive integers except the 11 numbers which are prime or powers of 2.

Not as elegant as the answer I look forward to seeing tomorrow - but it got the job done.

😀

@GWOMaths @threadreaderapp unroll