# Enigmatic Code

Programming Enigma Puzzles

## Enigma 1384: Power grid

From New Scientist #2544, 25th March 2006

I have drawn a grid of five rows of nine boxes and have placed in each box a digit so that each digit occurs its own number of times (e.g. 4 occurs four times). The digits in one row form a number which is a fourth power, and another row contains only two different digits and is divisible by 33. Two other rows between them contain all the digits and are both fifth powers.

In ascending order, what are the nine digits occupying the remaining row?

[enigma1384]

### One response to “Enigma 1384: Power grid”

1. Jim Randell 12 October 2013 at 9:26 am

This program looks for one arrangement of the number composed of only two different digits (as “anagrams” of this number won’t change the overall solution). It runs in 38ms.

```from collections import Counter
from itertools import combinations
from enigma import irange, intc, printf

# there are 45 boxes and 45 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9,
# so every digit is used in the solution.

digits = Counter(dict((d, int(d)) for d in '0123456789'))

# check the string provided doesn't exceed the digit count
def check(s):
return not any(v > digits[k] for (k, v) in Counter(s).items())

# find nth powers with 9 digits that don't exceed the digit count
def powers(n):
ps = list()
f = 1.0 / n
for i in irange(intc(100000000 ** f), int(999999999 ** f)):
s = str(i ** n)
if check(s): ps.append(s)
return ps

# we need to find 4th and 5th powers with 9 digits
p4s = powers(4)
p5s = powers(5)

# choose 2 of the 5th powers
for (p5a, p5b) in combinations(p5s, 2):
if not check(p5a + p5b): continue
if not all(d in p5a + p5b for d in '123456789'): continue
# choose the 4th powers
for p4 in p4s:
if not check(p4 + p5a + p5b): continue
# what digits are left?
ds = digits - Counter(p4 + p5a + p5b)
# choose two of them
for (d1, d2) in combinations(ds.keys(), 2):
if ds[d1] + ds[d2] < 9: continue
(d1, d2) = sorted((d1, d2), key=lambda x: ds[x])
# decide how many of the lesser digit to use
for n in irange(1, max(ds[d1], 9 - ds[d2])):
# will the resulting number be divisible by 3?
if (n * int(d1) + (9 - n) * int(d2)) % 3: continue
# and the positions it will occur in
for ps in combinations(irange(0, 8), n):
# create the number
n2 = ''.join((d1 if i in ps else d2) for i in irange(0, 8))
# check the disibility
if int(n2) % 33: continue
# and what's left?
r = ''.join(sorted((ds - Counter(n2)).elements()))

printf("r={r} [p4={p4} p5a={p5a} p5b={p5b} n2={n2}]")

# we don't need to find further possibilities for n2
break
```

Solution: The digits in the remaining row are: 3, 4, 5, 5, 8, 8, 8, 8, 8.

There are, in fact, 30 possibilities for the number composed of only two different digits, but the 4th and 5th powers are unique.

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