»  Solutions to puzzles in my VDARE.com monthly Diary

  August 2018


My August diary included the following brainteaser.

The arithmetic sequences a1, a2, a3, …, a20 and b1, b2, b3, …, b20 consist of 40 distinct positive integers, and a20 + b14 = 1000. Compute the least possible value for b20 + a14.


• Solution

First let's get a grip on our notation.

I shall henceforth refer to a1 as just a and b1 as just b.

The steps in the two series I'll refer to as h and k.

So now I can write the two series as

a, a+h, a+2h, a+3h, …, a+19h


b, b+k, b+2k, b+3k, …, b+19k

The thing given (datum), i.e. that a20 + b14 = 1000, translates into my notation as

a + 19h + b + 13k = 1000

Rearranging slightly:

a + b = 1000 − 19h − 13k

The thing sought (quaesitum) is the minimum possible value of a14 + b20; or, in my notation,

a + 13h + b + 19k

Substituting a + b from my rearrangement of the datum, that means I seek the minimum possible value of

1000 + 6(k − h)

[Note added later:   At this point I went astray. The logic that follows, while sound, assumes that h and k must be positive. In fact, as a reader pointed out, this is not required by the terms of the problem as stated. The ai and bi have to be positive, but the sequences might be de-scending.

If h and k are allowed to be negative, then 1000 + 6(k − h) can take values as low as 4 (i.e. when h = −1 and k − h = −167). I can't find a solution matching this value, but the next-lowest possibility 1000 + 6(k − h) = 10 does have solutions matching the terms of the problem:  a = 22, b = 3155, h = −1, k = −166.

My solution continues under the original but unjustified assumption that h and k are positive …]

Well, that's easy! The smallest possible value of that thing occurs when k is as small as possible and h is as large as possible. Since we are restricted to positive integers, the smallest possible value of k is 1. What is the largest possible value of h?

Given that number 1000, and given 20 as the number of terms in each sequence, 50 is a decent guess for the biggest possible h. Then a20 + b14 = a + b + 19h + 13k = a + b + 963. Then a + b would be 37, which sounds a bit high.

Stepping h up to 51 increases that 963 to 982, nicely closer to 1000. If we stepped it up again, to 52, the 982 would become 1001 and we'd need negative a + b to make the datum work — not allowed!

So h = 51 is the biggest possible value of h, and a + b is 18, from the datum (remember k is 1). We can then calculate b20 + a14. In my notation it's b + 19k + a + 13h. With a + b being 18 (because it's 1000 minus 982), h being 51, and k being 1, this comes out to 700.

That's the answer, and we're finished. Note as a cute sidebar point, however, that while the only thing we have deduced about a and b is that a + b = 18, we are not totally free with the individual values of a and b. Why not? Because the problem as stated tells us that the two sequences consist of 40 distinct positive integers.

Suppose for example we say that a = 17 and b = 1. Then a1 and b17 have the same value; they are both 17, violating the terms of the problem.

That should set off an alarm in the teased brain. Could it be that there are no allowable values of a and b with h = 51 and k = 1? Shall we have to go looking for other possibilities for h and k?

A moment's thought shows that we are in the clear. If a = 1 and b = 17, for example, then the b-sequence consists of the whole numbers from 17 to 36, none of which occurs in the a-sequence. So long as b is bigger than 9, we're in the clear. Phew!