Generalised solution to 1974 puzzle
Corey Plover
Allen L Truslove, Melbourne, VIC 3000
Corey.Plover@truslove.com.au
1.
Introduction
Page 23 of Actuary Australia, Volume 92 [1] issued a challenge to readers where every year
an uncle constructs integers using the digits 1974 in order and the mathematical symbols.
+ –   ^ !  ( ).
According to the author, the first age that cannot be solved is 113. This was improved by
various readers who managed to obtain all ages up to 150 [2].
This paper outlines a method that can construct every integer using an extension of factorials
called multifactorial products.
2.
Proof
Definition: The multifactorial product of n is the product of every kth integer from n to 1. For
example, 8!!! = 8  5  2 = 80:
if 0  n  k

1
n !! !  
n   n  k  !! ! if n  k
k times

k times

Lemma 1: All multiples of 2a can be constructed from a single integer, a ≥ 3.
Proof: With sufficient factorials, the multifactorial product contains only 2 non-trivial terms:
n !! !  nj  j   n2 
n  j times
Hence any product of integers can be converted to nested multifactorial product involving
only the first integer, so long as each integer is no greater than half of the accumulated
product. For a  3 and j sufficiently large, the following expression satisfies this condition:
a  2 2
 2 a  a 
ak k 
2 j 1 times
2 j times
 k   2ak 
2j
2 j times
Applying j squares roots to the above expression yields 2ak.
Corollary: If a = 2m, all even integers can be constructed by applying j square roots to:
2m  2  2 
 2 k  k 
2 j  m times
2 j times
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 k   2k 
2j
Lemma 2: All integers can be constructed from the pair of numbers, a ≥ 3 and b ≥ 3.
Proof: Any non-zero integer k can be expressed in the following form:
2abk
2ab
2abk and 2ab can each be constructing from the numbers a and b respectively. Zero can also
be constructed by constructing 2ab from each of a and b separately and subtracting.
Lemma 3: All non-zero integers can be constructed from a = 2m ≥ 3 and any other number, b.
Proof:

Case b  3: Special case of Lemma 2.

Case b = 2: From the corollary of Lemma 1 all even numbers can be constructed from
a = 2m. Division by b = 2 can then yield any non-zero integer k.

Case b = 1: From the corollary of Lemma 1 all even numbers can be constructed from
a = 2m. Adding or multiplying by b = 1 can then yield any odd or even number
respectively.

Case b = 0: From the corollary of Lemma 1 all even numbers can be constructed from
a = 2m. Adding b! = 1 or b = 0 can then yield any odd or even number respectively.
Conjecture 1: All integers can be constructed from any ordered 4 digit set {a,b,c,d}, except if
the set consists solely of 0’s and 1’s.
Proof: By a few generalisations (like when a = 2 or 4, form ab) and enumeration of remaining
cases, most sets of 2 digits and all sets of three digits (except those consisting solely of 0’s
and 1’s) can form a number a = 2m.
One of the sets {a,b,c} or {b,c,d} will therefore allow a number 2m to be formed, unless all
digits are 0’s or 1’s. The proof then follows from application of Lemma 3.
Note that the sets {1,0,1,0}, {1,0,1,1}, {1,1,1,0} and {1,1,1,1} can be partitioned into 10’s
and 11’s and Lemma 2 applied.
Conjecture 2: All integers can be constructed from any ordered 5 or more digit set
Proof: From the previous conjecture, 4 digits suffices unless all digits are 0’s or 1’s. If this is
the case, the first 4 digits can be combined to form a! + b! + c! + d! = 4 and Lemma 3 applied.
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3.
Example
Here we construct the number k = 150 from the set of digits {1,9,7,4}.
3.1. Formed using a  3 and b  3
Let us assume a = 19 and b = 74 (although many other partitions are possible). We construct
2abk = 421,800 from a and 2ab = 2,812 from b using the product chains:
421,800  19  2  2 19  74  74 150 150
2,812  74  2 19
Simpler product chains are possible that satisfy the accumulated product rule from Lemma 1
(i.e. 421,800 = 19  8  37  75) but may not exist in general, especially when a, b and/or k
are prime. Combining our original product chains yields:
150 
(19 !! !) !! !) !! !) !! ! ) !! ! )
!! !
)
!! !
17 times 36 times 57 times 1,370 times 106,782 times 7,907,194 times 1,186,101,450 times
)
(74 !! !) !! ! )
72 times 129 times
Using 1,194,117,107 exclamation points in total!
3.2. Formed using a = 2m and b
Let us assume b = 1 and a = (9 – 7)4 = 16 (although again, many other partitions are possible).
We construct k = 150 from the number a using the product chain:
150 
16  2  2  2  2  75  75 
 75
8 times
This factorial expression is too long to show here. We simplify the product chain slightly
while still continuing to satisfy the accumulated product rule:
150 
16  3  5  5  75  75  75
Combining the resulting factorial expressions with b = 1 gives:
150  1
9  7
4
!! !) !! !) !! ! ) !! ! ) !! ! )
!! !
13 times 43 times 235 times 1,125 times 89,925 times 6,749,925 times
)
Using only 6,821,266 exclamation points in total!
4.
References
[1]
Stephen Woods, “Two Ducks Swimming”, Actuary Australia 92 (2004), 22-23.
[2]
Stephen Woods, “Two Ducks Swimming”, Actuary Australia 99 (2005), 22-23.
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