Sunday Times Teaser 3049 – Plantation

by BRG on February 26, 2021

by Andrew Skidmore

Published Sunday February 28 2021 (link)

Jed farms a large, flat, square area of land. He has planted trees at the corners of the plot and all the way round the perimeter; they are an equal whole number of yards apart.

The number of trees is in fact equal to the total number of acres (1 acre is 4840 square yards). If I told you an even digit in the number of trees you should be able to work out how far apart the trees are.

How many yards are there between adjacent trees?

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Brian Gladman permalink


from collections import defaultdict

# map even digits in the number of trees to solutions
ed2sn = defaultdict(list)

# With T as the number of trees (and the land area in acres) the
# area side length in yards is sqrt(4840.T) = 22.sqrt(10.T);  to
# make the side length integer requires that T = 10.M^2 for some
# integer M, giving a side length of 220.M; the tree spacing can 
# now be determined as 880.M / 10.M^2 = 88 / M; M also has to be
# even (=2.K) since T has to be a multiple of four

for K in range(1, 45):
  spacing, r = divmod(44, K)
  if not r:
    trees = str(40 * K ** 2)
    print(f"{trees:5} trees {spacing:2} yards apart")

    # store solutions indexed on their even digits
    for d in '02468':
      if d in trees:
        ed2sn[d].append((spacing, trees))

print()
# find a solution that has a unique even digit
for d, l_sn in ed2sn.items():
  (s, n), *r = l_sn
  if not r:
    print(f"Spacing: {s} yards ({n} trees)")

from collections import defaultdict

# map even digits in the number of trees to solutions

ed2sn = defaultdict(list)

# With T as the number of trees (and the land area in acres) the

# area side length in yards is sqrt(4840.T) = 22.sqrt(10.T); to

# make the side length integer requires that T = 10.M^2 for some

# integer M, giving a side length of 220.M; the tree spacing can

# now be determined as 880.M / 10.M^2 = 88 / M; M also has to be

# even (=2.K) since T has to be a multiple of four

for K in range(1, 45):

spacing, r = divmod(44, K)

if not r:

trees = str(40 * K ** 2)

print(f"{trees:5} trees {spacing:2} yards apart")

# store solutions indexed on their even digits

for d in '02468':

if d in trees:

ed2sn[d].append((spacing, trees))

print()

# find a solution that has a unique even digit

for d, l_sn in ed2sn.items():

(s, n), *r = l_sn

if not r:

print(f"Spacing: {s} yards ({n} trees)")

Frits permalink

@Brian,

Did you consider:

Python

from number_theory import divisors for M in [d for d in divisors(88) if d % 2 == 0]:

1
2
3
4
5

from number_theory import divisors

for M in [d for d in divisors(88) if d % 2 == 0]:

Thanks for introducing the line 29/30 approach. I would have done a len(..) == 1 statement.

Reply
- Brian Gladman permalink
  
  @Frits Yes, I was aware of that approach. But I try to avoid the need for
  non-standard library imports unless they are really needed in order to
  provide published code that runs ‘out of the box’ without the need for
  other files. The problem with this teaser from a coding perspective
  is that it can be easily analysed it to the point where its not even
  worth coding!
  
  Reply

Erling Torkildsen permalink


# Map the number of trees to the tree spacing
T2yrds, tt = dict(), ""

# With T as the number of trees (and the land 
# area in acres), n as the tree spacing and A
# as the land area in square yards:
#
# A = (n.T/4)^2 = 4840.T ==> T = 40.(44/n)^2
 
for n in range(1, 45):
  if 44 % n == 0:
    T = 40 * (44 // n) ** 2
    T2yrds[T] = n
    tt += str(T)

# Find the 'even' which occurs once
for e in '02468':
  if tt.count(e) == 1:
    for T in T2yrds:
      if e in str(T):
        print(f"{T2yrds[T]} yards ({T} trees).")

# Map the number of trees to the tree spacing

T2yrds, tt = dict(), ""

# With T as the number of trees (and the land

# area in acres), n as the tree spacing and A

# as the land area in square yards:

# A = (n.T/4)^2 = 4840.T ==> T = 40.(44/n)^2

for n in range(1, 45):

if 44 % n == 0:

T = 40 * (44 // n) ** 2

T2yrds[T] = n

tt += str(T)

# Find the 'even' which occurs once

for e in '02468':

if tt.count(e) == 1:

for T in T2yrds:

if e in str(T):

print(f"{T2yrds[T]} yards ({T} trees).")

Frits permalink

@Erling, this version is easier to read but surely (44 % n) equals 44 for n > 44

Reply
- Brian Gladman permalink
  
  @Frits. That was my fault with some faulty editing, corrected now.
  
  Reply

Sunday Times Teaser 3049 – Plantation

by Andrew Skidmore

Published Sunday February 28 2021 (link)

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