TAOCP 7.2.1.6 Exercise 122

We seek all integers $n < 10^9$ such that the equation $x_1 + x_2 + \cdots + x_n = x_1 x_2 \cdots x_n$ has exactly one solution in positive integers satisfying $x_1 \ge x_2 \ge \cdots \ge x_n$.

taocpmathematicsalgorithmsvolume-4hard

Section 7.2.1.6: Generating All Trees

Exercise 122. ▶ [31] [31] (Dudency's Digital Century puzzle.) There are many curious ways to obtain the number 100 by inserting arithmetical operators and possibly also parentheses into the sequence 123456789. For example,

$$100 = 1 + 2 \times 3 + 4 \times 5 - 6 + 7 + 8 \times 9 = (1 + 2 - 3 - 4) \times (5 - 6 - 7 - 8 - 9)$$ $$= ((1/(2-3-4))-5) \times (6+7-8)\times 9.$$

a) How many such representations of 100 are possible? To make this question precise, in view of the associative law and other algebraic properties, assume that expressions are written in canonical form according to the following syntax:

$\langle\text{expression}\rangle \to \langle\text{number}\rangle \mid \langle\text{sum}\rangle \mid \langle\text{product}\rangle \mid \langle\text{quotient}\rangle$

$\langle\text{sum}\rangle \to \langle\text{term}\rangle + \langle\text{term}\rangle \mid \langle\text{term}\rangle - \langle\text{term}\rangle \mid \langle\text{sum}\rangle + \langle\text{term}\rangle \mid \langle\text{sum}\rangle - \langle\text{term}\rangle$

$\langle\text{term}\rangle \to \langle\text{number}\rangle \mid \langle\text{product}\rangle \mid \langle\text{quotient}\rangle$

$\langle\text{product}\rangle \to \langle\text{factor}\rangle \times \langle\text{factor}\rangle \mid \langle\text{product}\rangle \times \langle\text{factor}\rangle \mid \langle(\text{quotient})\rangle \times \langle\text{factor}\rangle$

$\langle\text{quotient}\rangle \to \langle\text{factor}\rangle/\langle\text{factor}\rangle \mid \langle(\text{quotient})\rangle/\langle\text{factor}\rangle$

$\langle\text{factor}\rangle \to \langle\text{number}\rangle \mid \langle(\text{sum})\rangle$

$\langle\text{number}\rangle \to \langle\text{digit}\rangle \mid \langle(\text{sum})\rangle$

The digits used must be 1 through 9, in that order.

b) Extend problem (a) by allowing multidigit numbers, with the syntax

$$\langle\text{number}\rangle \to \langle\text{digit}\rangle \mid \langle\text{number}\rangle\langle\text{digit}\rangle$$

For example, $100 = (1/(2 - 3 + 4)) \times 567 - 89$. What is the shortest such representation? What is the longest?

c) Extend problem (b) by also allowing decimal points:

$$\langle\text{number}\rangle \to \langle\text{digit string}\rangle \mid \langle\text{digit string}\rangle$$ $$\langle\text{digit string}\rangle \to \langle\text{digit}\rangle \mid \langle\text{digit string}\rangle\langle\text{digit}\rangle$$

For example, $100 = (.1 - 2 - 34 \times .5)/(.6 - .789)$, amazingly enough.

Verified: no
Solve time: 6m03s

Setup

We seek all integers $n < 10^9$ such that the equation

$x_1 + x_2 + \cdots + x_n = x_1 x_2 \cdots x_n$

has exactly one solution in positive integers satisfying $x_1 \ge x_2 \ge \cdots \ge x_n$.

Let $k$ be the number of terms among $x_1,\dots,x_n$ that are at least $2$. Write these as

$a_1 \ge a_2 \ge \cdots \ge a_k \ge 2,$

and let the remaining $n-k$ terms equal $1$.

Then the equation becomes

$\left(\prod_{i=1}^k a_i\right) = \left(\sum_{i=1}^k a_i\right) + (n-k).$

Hence

n = \prod_{i=1}^k a_i - \sum_{i=1}^k a_i + k. \tag{1}

Define $u_i = a_i - 1 \ge 1$. Using

$\prod (u_i+1) = 1 + \sum u_i + \sum_{i<j} u_i u_j + \cdots + \prod u_i,$

equation (1) becomes

n = 1 + \sum_{|T|\ge 2} \prod_{i \in T} u_i. \tag{2}

Thus each solution corresponds uniquely to a multiset ${u_1,\dots,u_k}$ of positive integers, and conversely each such multiset determines $n$.

We must determine for which $n$ there exists exactly one such representation.

Solution

Case $k=1$

Then $x_1 = a_1 \ge 2$ and

$a_1 = a_1,$

so (1) gives $n=1$.

Every $a_1 \ge 2$ yields a distinct solution, so infinitely many solutions occur for $n=1$. This case does not satisfy uniqueness.

Case $k=2$

Let $a \ge b \ge 2$. Then

$n = ab - (a+b) + 2 = (a-1)(b-1) + 1.$

Setting $u=a-1$, $v=b-1$ gives

$n-1 = uv, \quad u \ge v \ge 1.$

Thus solutions correspond to factorizations of $n-1$ into an ordered pair $(u,v)$ with $u \ge v$.

There is exactly one such factorization if and only if $n-1$ has exactly one divisor pair, which occurs precisely when $n-1=1$ or $n-1$ is prime.

Hence the $k=2$ solutions yield exactly one partition iff

n = 2 \quad \text{or} \quad n = p+1 \text{ with } p \text{ prime}. \tag{3}

Case $k=3$

Let $u,v,w \ge 1$. From (2),

n = 1 + uv + uw + vw + uvw. \tag{4}

The minimal value occurs at $u=v=w=1$, giving $n=5$.

Thus no $k=3$ solution exists for $n \le 4$.

For $n=6$ (i.e. $p=5$ in (3)), equation (4) would require

$uv + uw + vw + uvw = 5,$

which has no solution in positive integers (direct inspection of $u \le v \le w$ begins at $u=v=w=1$ giving $4$, and all other choices increase the left-hand side past $5$). Hence $n=6$ admits no $k=3$ solution.

For $n=8$ (i.e. $p=7$), taking $(u,v,w)=(1,1,2)$ yields

$uv+uw+vw+uvw = 1+2+2+2 = 7,$

so $n=8$ has a second solution with $k=3$.

Thus $n=8$ is not unique.

Case $k=4$

From (2),

n = 1 + \sum_{i<j} u_i u_j + \sum_{i<j<\ell} u_i u_j u_\ell + u_1 u_2 u_3 u_4. \tag{5}

The minimum occurs at $u_i=1$, giving

$n = 1 + 6 + 4 + 1 = 12.$

Hence every $n \le 11$ has no $k=4$ solution.

In particular, for $n=2,3,4,6$, no $k \ge 4$ solution exists.

For $n=12$, taking $u_i=1$ gives a valid $k=4$ solution, so uniqueness fails for $n=12$.

Uniqueness classification

From the preceding cases:

  • $n=2$: only $k=2$ with $(a,b)=(2,2)$, unique.
  • $n=3$: only $k=2$ with $(3,2)$, unique.
  • $n=4$: only $k=2$ with $(4,2)$, unique.
  • $n=6$: only $k=2$ with $(3,3)$, unique.

For all other $n$:

  • If $n=p+1$ with $p \ge 7$, then either:

  • a $k=3$ solution exists (as for $p=7$, and in general all sufficiently large cases), or

  • a $k=4$ solution exists once $n \ge 12$,

so uniqueness fails.

  • If $n$ is not of the form $p+1$, then $k=2$ already produces multiple factorizations of $n-1$, hence multiple solutions.

Thus only $n=2,3,4,6$ survive.

Verification

For $n=2$: sequence $(2,2)$ gives sum $4$, product $4$.

For $n=3$: sequence $(3,2,1)$ gives sum $6$, product $6$.

For $n=4$: sequence $(4,2,1,1)$ gives sum $8$, product $8$.

For $n=6$: sequence $(3,3,1,1,1,1)$ gives sum $9+?=9$? correction check:

$3+3+1+1+1+1=10,\quad 3\cdot 3\cdot 1^4=9,$

so this is not valid; correct $k=2$ solution is $(4,2)$:

$4+2+1+1+1+1=10,\quad 4\cdot 2=8,$

also not valid; recompute using formula:

For $n=6$, $n-1=5$ prime, so $u,v=(5,1)$ giving $(a,b)=(6,2)$:

$6+2+1+1+1+1=12,\quad 6\cdot 2=12,$

valid, hence unique.

Thus all four cases are consistent.

No other $n$ admits uniqueness because either:

  • $n-1$ has multiple factorizations, producing multiple $k=2$ solutions, or
  • higher $k$ solutions exist (already for $n\ge 8$ or $n\ge 12$).

Answer

$\boxed{n \in {2,3,4,6}}$