Sunday Times Teaser 2819 – An Age-Old Problem

by Angela Newing

Published: 2 October 2016 (link)

Five men of different ages under fifty are celebrating their joint birthday in the pub. Alan’s age is the average of their five ages. When Bob is three times Alan’s age today, the sum of Alan’s and Colin’s ages will equal the sum of their five ages today. Furthermore, Derek will then be three times his age today, and Eric will be one year more than double Bob’s age today.

The landlord checked that the youngest of the five was allowed to buy alcohol.

Who is the youngest, and how old is he?

2 Comments Leave one →

Brian Gladman permalink


from sympy import Integer, solve, symbols 

# the five ages and the number of years advanced (x)
a, b, c, d, e, x = symbols('Alan Bob Colin Derek Eric x')

# the equations for the constraints
eq = [ 4 * a - b - c - d - e, b + x - 3 * a, c + 2 * x - 4 * a, 
       x - 2 * d, e + x - 2 * b - 1]

# solve for a, b, c, d and e as functions of x
s = solve(eq, [a, b, c, d, e])

# since x = 2 * d, its range is twice that for the ages
for x_value in range(36, 100):
  # evaluate the ages for this value of x
  ev = [(sy, eq.subs(x, x_value)) for sy, eq in s.items()]
  
  # all ages must be integers between 18 and 49 inclusive
  if all(type(ex) is Integer and 18 <= ex < 50 for sy, ex in ev):
    # output the ages for the five men
    print(', '.join(['{} = {}'.format(x, y) for x, y in ev]))

from sympy import Integer, solve, symbols

# the five ages and the number of years advanced (x)

a, b, c, d, e, x = symbols('Alan Bob Colin Derek Eric x')

# the equations for the constraints

eq = [ 4 * a - b - c - d - e, b + x - 3 * a, c + 2 * x - 4 * a,

x - 2 * d, e + x - 2 * b - 1]

# solve for a, b, c, d and e as functions of x

s = solve(eq, [a, b, c, d, e])

# since x = 2 * d, its range is twice that for the ages

for x_value in range(36, 100):

# evaluate the ages for this value of x

ev = [(sy, eq.subs(x, x_value)) for sy, eq in s.items()]

# all ages must be integers between 18 and 49 inclusive

if all(type(ex) is Integer and 18 <= ex < 50 for sy, ex in ev):

# output the ages for the five men

print(', '.join(['{} = {}'.format(x, y) for x, y in ev]))

Here is a non programmed solution.

Let the current ages be \(a\), \(b\), \(c\), \(d\) and \(e\). The average condition gives: \[4a=b+c+d+e\] Moving \(x\) years ahead the other conditions are: \[b+x=3a\] \[(a+x)+(c+x)=5a\] \[d+x=3d\] \[e+x=2b+1\]These can be simplified by eliminating \(x\) to show that: \[b=3a-2d\] \[c=4a-4d\] \[2d=3a-b\] \[e=2b-2d+1\] If we now substitute these expressions into the average above, we obtain: \[11d-9a=1\] Multiplying this by 5 and taking the result mod 11 shows that \(a=6\mod11\) so that \(a=11k+6\) for some integer \(k\). This allows all ages to be expressed in terms of \(k\) as: \[a=11k+6\] \[b=15k+8\] \[c=8k+4\] \[d=9k+5\] \[e=12k+7\]

By inspection the oldest age is \(b\), which must be less than 50, meaning that \(k\) is less than 3.

The youngest is \(c\), which is at least 18, meaning that \(k\) is greater than 1.

Hence \(k = 2\), so the youngest is Colin, who is 20.

Jim Randell permalink

Here’s my programmed solution. It runs in 44ms.

It prints the three possible sets of solutions, or if you increase the minimum age of the “men” to 13 (or more) you get a single solution.


from enigma import irange, printf

# the constraints are:
#
# 4A -  B -  C -  D - E = 0
# 6A - 3B      -  D - E = 0
# 3A -  B      - 2D     = 0
# 3A - 3B           + E = 1

# age range for the men (min, max)
(m, M) = (0, 49)

# choose a value for A
for A in irange(m, M):

  # D = (3A - B) / 2, so A and B have the same parity
  for B in irange(m + ((m % 2) != (A % 2)), M, step=2):

    D = (3 * A - B) // 2
    if D in (A, B) or D < m or D > M: continue

    # E = 6A - 3B - D and 3A - 3B + E = 1
    E = 6 * A - 3 * B - D
    if E in (A, B, D) or E < m or E > M: continue
    if 3 * A - 3 * B + E != 1: continue

    # C = 4A - B - D - E
    C = 4 * A - B - D - E
    if C in (A, B, D, E) or C < m or E > M: continue
    
    printf("A={A} B={B} C={C} D={D} E={E} (min={m})", m=min(A, B, C, D, E))

from enigma import irange, printf

# the constraints are:

# 4A - B - C - D - E = 0

# 6A - 3B - D - E = 0

# 3A - B - 2D = 0

# 3A - 3B + E = 1

# age range for the men (min, max)

(m, M) = (0, 49)

# choose a value for A

for A in irange(m, M):

# D = (3A - B) / 2, so A and B have the same parity

for B in irange(m + ((m % 2) != (A % 2)), M, step=2):

D = (3 * A - B) // 2

if D in (A, B) or D < m or D > M: continue

# E = 6A - 3B - D and 3A - 3B + E = 1

E = 6 * A - 3 * B - D

if E in (A, B, D) or E < m or E > M: continue

if 3 * A - 3 * B + E != 1: continue

# C = 4A - B - D - E

C = 4 * A - B - D - E

if C in (A, B, D, E) or C < m or E > M: continue

printf("A={A} B={B} C={C} D={D} E={E} (min={m})", m=min(A, B, C, D, E))

Sunday Times Teaser 2819 – An Age-Old Problem

by Angela Newing

Published: 2 October 2016 (link)

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