Sunday Times Teaser 2874 – An Age-Old Progression

by Angela Newing

Published October 22 2017 (link)

It is my birthday today — the same day as two of my younger relatives, Betty and Charles. I commented that the digits involved in our three ages are all different. Betty noted that the square of her age is equal to my age multiplied by Charles’s age. Then Charles added that on one of our birthdays in the next ten years the sum of our ages will be one hundred.

What are our three ages today?

4 Comments Leave one →

Brian Gladman permalink


# since a > b and a + b + c < 100, b is less than 50
for b in range(1, 50):
  # find divisors of b squared (a.c =  b^2)
  for c in range(1, b):
    a, r = divmod(b * b, c)
    if not r:
      # compute the number of years (0 < n < 11)
      # to when the sum of their ages is 100 
      n, r = divmod(100 - (a + b + c), 3)
      if c > 1 and not r and 0 < n < 11:
        ss = ''.join(str(x) for x in (a, b, c))
        if len(ss) == len(set(ss)):
          print(f'Angela is {a}, Betty is {b} and Charles is {c} (n = {n})')

# since a > b and a + b + c < 100, b is less than 50

for b in range(1, 50):

# find divisors of b squared (a.c = b^2)

for c in range(1, b):

a, r = divmod(b * b, c)

if not r:

# compute the number of years (0 < n < 11)

# to when the sum of their ages is 100

n, r = divmod(100 - (a + b + c), 3)

if c > 1 and not r and 0 < n < 11:

ss = ''.join(str(x) for x in (a, b, c))

if len(ss) == len(set(ss)):

print(f'Angela is {a}, Betty is {b} and Charles is {c} (n = {n})')

Trevor Kirsten permalink


# c < b < a < 100, b^2 = ac (:. b is not prime)
# a + b + c = 100 - 3n = x, 0 < n < 11 (:. 4 < b < 46)
# this reduces to a = (x - b + (x * x - 2 * b * x - 3 * b * b) ** 0.5) / 2
 
for b in range(4, 47):
  for x in range(70, 100, 3):
    if x * x >= 2 * b * x + 3 * b * b:
      a = (x - b + (x * x - 2 * b * x - 3 * b * b) ** 0.5) / 2
      if a == a // 1:
        c = b * b / a
        if c == c // 1:
          print(a, b, c, x)

# c < b < a < 100, b^2 = ac (:. b is not prime)

# a + b + c = 100 - 3n = x, 0 < n < 11 (:. 4 < b < 46)

# this reduces to a = (x - b + (x * x - 2 * b * x - 3 * b * b) ** 0.5) / 2

for b in range(4, 47):

for x in range(70, 100, 3):

if x * x >= 2 * b * x + 3 * b * b:

a = (x - b + (x * x - 2 * b * x - 3 * b * b) ** 0.5) / 2

if a == a // 1:

c = b * b / a

if c == c // 1:

print(a, b, c, x)

Brian, my code for checking no repeat digits in a,b & c is cumbersome & longer than my “basic” code above so I omitted it – please help!

Brian Gladman permalink

Hi Trevor, thanks for your post. To check for repeated digits, it is useful to have a, b and c as integers so I modified your code accordingly in order to add this check:


for b in range(4, 47):
  for x in range(70, 100, 3):
    t = x * x - b * (2 * x + 3 * b)
    if t > 0:
      r = int(t ** 0.5)
      if t == r * r:
        a = (x - b + r) // 2
        c = b * b // a
        s = str(a) + str(b) + str(c)
        if len(s) == len(set(s)):
          print(a, b, c, x)

for b in range(4, 47):

for x in range(70, 100, 3):

t = x * x - b * (2 * x + 3 * b)

if t > 0:

r = int(t ** 0.5)

if t == r * r:

a = (x - b + r) // 2

c = b * b // a

s = str(a) + str(b) + str(c)

if len(s) == len(set(s)):

print(a, b, c, x)

Trevor Kirsten permalink

Hi Brian. I think I get it now. By converting variables to integers gets rid of decimal points & zeros after the points so that the set function doesn’t get confused & treat them as repeats.
Many thanks.

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Sunday Times Teaser 2874 – An Age-Old Progression

by Angela Newing

Published October 22 2017 (link)

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