# »  Solutions to puzzles in my VDARE.com monthly Diary

## August 2018

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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.

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• 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

and

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!