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1 Answer

100×2ⁿ=25×4×2ⁿ=25×2ⁿ⁺².

Let p and q be two positive integers. The sum of all the positive integers up to each of these is p(p+1)/2 and q(q+1)/2. That is, the sum of all integer x for q<x≤p. The difference between these is the sum of all the positive integers between p and q. So if p>q, p(p+1)/2-q(q+1)/2=25×2ⁿ⁺², p²+p-q²-q=25×2ⁿ⁺³, p²-q²+p-q=25×2ⁿ⁺³, (p-q)(p+q)+(p-q)=25×2ⁿ⁺³, (p-q)(p+q+1)=25×2ⁿ⁺³.

So, let p+q+1=25, therefore p+q=24; and p-q=2ⁿ⁺³.

Add these two equations: 2p=24+2ⁿ⁺³, p=12+2ⁿ⁺² and q=24-p=12-2ⁿ⁺².

Since q>0, 12-2ⁿ⁺²>0, 3-2ⁿ>0, 2ⁿ<3, so n≤1. Because p and q are positive integers, n≥-2. q<1000, so 12-2ⁿ⁺²<1000, 3-2ⁿ<250, 2ⁿ>-247, which is true for all n. So the range is -2≤n≤1, making n ∈ {-2 -1 0 1} giving (p,q,n)={ (13,11,-2) (14,10,-1) (16,8,0) (20,4,1) }.

For n=1, q=12-8=4 and p=12+8=20.

4×5/2=10, 20×21/2=210; 210-10=200=100×2.

Therefore 5+6+7+...+19+20=200.

Now let p+q+1=2ⁿ⁺³ and p-q=25. 

p+q=2ⁿ⁺³-1, p-q=25. Add these: 2p=2ⁿ⁺³+24, p=2ⁿ⁺²+12, q=2ⁿ⁺²-13.

q>0, so 2ⁿ⁺²>13, n+2≥4, n≥2. But p<1000, 2ⁿ⁺²+12<1000, 2ⁿ⁺²<988, n≤7.

Therefore, 2≤n≤7.

For n=7, p=524 and q=499.

524×525/2=137550 and 499×500/2=124750, 137550-124750=12800.

Therefore: 500+501+...+523+524=12800.

Continued in comment...

by Top Rated User (762k points)

Note that there are many more solutions to this problem and all can be found by applying the method shown and using all values of n in the discovered range.

Also, (p-q)(p+q+1)=5×(5×2ⁿ⁺³) and solving for simple systems (p-q=5 and p+q=5×2ⁿ⁺³-1) and (p-q=5×2ⁿ⁺³ and p+q=4) produce more solutions for (p,q,n), including:

(642,637,5): 638+639+640+641+642=3200;

(7,2,-2): 3+4+5+6+7=25.

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