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A recreational math problem, dubbed "Insert and Add", asks: What is the least integer m that requires no less than n insertions of plus signs so that, after performing the addition(s), we arrive at a single digit? (See the last page here: http://orion.math.iastate.edu/butler/papers/16_03_insert_and_add.pdf)

It is similar to finding the additive persistence of n, but instead of merely counting the number of digital sums required to arrive at a single digit it counts the minimum number of plus signs inserted during that process.

10 is the smallest number that requires one plus sign: 1+0=1. 19 is the smallest to require two: 1+9=10 -> 1+0=1. 118 is the smallest to require three: 1+1+8=10 -> 1+0=1; alternatively we can try 1+18=19 -> 1+9=10 -> 1+0=1; and finally we can try 11+8=19 -> 1+9=10 -> 1+0=1.

3187, and 3014173 are the next two numbers in the sequence.

Now observe that all of these numbers (10, 19, 118, 3187, 3014173) have a digital root of 1.

Is it obvious that all future terms in this sequence will have digital root 1?

The sequence is https://oeis.org/A293929.

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  • $\begingroup$ jnthn - I have deleted my answer because there is a flaw in the argument. The series for m and m' need not stay 'in sync' just 1 apart. E.g. 3981 can be reduced to 39 + 81 = 119 and then to 11 + 9 =20. However, 3982 with the same insertions becomes 120 and then 12, which is 8 less rather than 1 more. $\endgroup$ – S. Dolan Nov 30 '17 at 23:38
  • $\begingroup$ Note that the digital root is preserved when you insert plus signs. I want to claim that if a proposed $a(n)$ has digital root greater than $1$ you can reduce the units digit by $1$ and get a smaller number that takes the same number of plus signs. If the units digit is $0$ you can reduce the digit that gets added to the units in the best solution. I can't make the argument quite work, though. $\endgroup$ – Ross Millikan Dec 2 '17 at 15:26
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I am terrible at proofs, so I won't be surprised when someone points out a glaring hole in this, but what about:

Assume the terms $a(1)$ to $a(n)$ all have digital root $1$, but $a(n + 1) = x$ doesn't. Increment $a(n)$ by one until we reach $x$.

Insert one plus sign into $x$ in the optimal way that guarantees the result of the addition, $y$, requires exactly $n$ more insertions of a plus sign to arrive at a single digit.

Because $y$ requires $n$ insertions it cannot be less than $a(n)$, otherwise we would have found $y$ before $a(n)$. Because $x$ has digital root greater than $1$, $y$ cannot equal $a(n)$. So now $y$ must be in the range $a(n) < y < x$, but we already checked these before arriving at $x$, so no such number $y$ can exist, therefore no such $x$ can exist. Clearly, $a(n + 1)$ cannot have digital root $0$.

We have shown that no $a(n + 1) = x$ with digital root $0$, or $2$ through $9$ can exist, therefore $a(n + 1)$ must have digital root $1$.

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  • $\begingroup$ The first sentence of the last paragraph can be deleted. The claim that $a(n) \lt y \lt x$ is correct, but this is not a problem because $y$ requires only $n$ plus signs, not $n+1$ $\endgroup$ – Ross Millikan Dec 1 '17 at 22:09
  • $\begingroup$ @Ross Millikan So I should remove: "This means y must be in the range a(n−1) < y < x."? $\endgroup$ – jnthn Dec 2 '17 at 0:36
  • $\begingroup$ I would, but looking again I think it is close. Once you say $y$ takes $n$ insertions you know it is greater than or equal to $a(n)$. If you want to say that explicitly, that is fine. $\endgroup$ – Ross Millikan Dec 2 '17 at 3:50
  • $\begingroup$ @Ross Millikan Done. Thank you, Sir, for the feedback. $\endgroup$ – jnthn Dec 2 '17 at 14:59

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