## August 2018

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My August diary included the following brainteaser.

The arithmetic sequencesa_{1},a_{2},a_{3}, …,a_{20}andb_{1},b_{2},b_{3}, …,b_{20}consist of 40 distinct positive integers, anda_{20}+b_{14}= 1000. Compute the least possible value forb_{20}+a_{14}.

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**• Solution**

First let's get a grip on our notation.

I shall henceforth refer to *a*_{1} as just *a* and *b*_{1} 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 *a*_{20} + *b*_{14} = 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 *a*_{14} + *b*_{20}; 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

*a*and

_{i}*b*have to be positive, but the sequences might be

_{i}*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
*a*_{20} + *b*_{14} = *a* + *b* + 19*h* + 13*k* =
*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 *b*_{20} + *a*_{14}. In my notation it's
*b* + 19*k* + *a* + 13*h*. 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 *a*_{1} and
*b*_{17} 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!