Sunday Times Brain Teaser 1023 – A Mixed Maths Class

by BRG on March 11, 2024

by Rachel Blunt

Published 7th March 1982 (see link)

My mathematics class consists of six boys and six girls. In their annual examination each was awarded integral an mark out of 100.

Disappointingly no boy received a distinction (over 80) but all the boys managed over 40. The lowest mark in the class was 36.

Upon listing the boys’ marks I noticed that all their marks were different prime numbers and that their average was an even number. Further three of the boys’ marks formed an arithmetic progression, and the other three another arithmetic progression.

Turning, my, attention to the girls I found that their marks were all different. There was little overall difference in the performance of the
sexes, the total of the girls’ marks being just one more than the total of the boys’. Three of the girls’ marks formed one geometric progression, and
the other three formed another geometric progression with the same ratio as the first one.

Finally when listing the results in numerical order I was pleased to see that Annie (who did so badly last year) had come seventh in the class.

What were the top six marks (in descending order)?

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BRG permalink


from itertools import combinations, product

# primes between 40 and 80
pb = set(n for n in range(41, 80, 2) if all(n % p for p in (3, 5, 7)))

# return the square root of <n> if it is an integer else None 
is_sq = lambda n: r if (r := round(n ** (1 / 2))) ** 2 == n else None

# collect triples in arithmetic progression for the boys
bps = [{a, b, c} for a, b in combinations(pb, 2) if (c := 2 * b - a) in pb]
# now find pairs of triples that don't share any values 
brs = []
for ap, bp in combinations(bps, 2):
  if not ap & bp and not sum(ap | bp) % 2:
    brs.append(tuple(sorted(ap | bp)))
    
# collect triples in geometric progression for the girls
gps = [(a, b, c) for a, c in combinations(range(36, 100), 2)
                        if (b := is_sq(a * c)) and a < b < c]
# now find pairs of triples that don't share any values 
grs = []
for ap, bp in combinations(gps, 2):
  if not set(ap).intersection(bp) and ap[2] * bp[0] == bp[2] * ap[0]:
    grs.append(tuple(sorted(ap + bp)))

# consider all combinations of the boys and girls scores ...
for br, gr in product(brs, grs):
  # ... in which the girls total marks are one more than the boys
  if sum(br) + 1 == sum(gr):
    # all marks in top to bottom order
    mks = sorted(br + gr, reverse=True)
    # a girl is in seventh position
    if mks[6] in gr:
      f, s, l = ', '.join(str(m) for m in mks[:6]).rpartition(', ')
      print(f"The top six marks are {' and '. join((f, l))}.")

from itertools import combinations, product

# primes between 40 and 80

pb = set(n for n in range(41, 80, 2) if all(n % p for p in (3, 5, 7)))

# return the square root of <n> if it is an integer else None

is_sq = lambda n: r if (r := round(n ** (1 / 2))) ** 2 == n else None

# collect triples in arithmetic progression for the boys

bps = [{a, b, c} for a, b in combinations(pb, 2) if (c := 2 * b - a) in pb]

# now find pairs of triples that don't share any values

brs = []

for ap, bp in combinations(bps, 2):

if not ap & bp and not sum(ap | bp) % 2:

brs.append(tuple(sorted(ap | bp)))

# collect triples in geometric progression for the girls

gps = [(a, b, c) for a, c in combinations(range(36, 100), 2)

if (b := is_sq(a * c)) and a < b < c]

# now find pairs of triples that don't share any values

grs = []

for ap, bp in combinations(gps, 2):

if not set(ap).intersection(bp) and ap[2] * bp[0] == bp[2] * ap[0]:

grs.append(tuple(sorted(ap + bp)))

# consider all combinations of the boys and girls scores ...

for br, gr in product(brs, grs):

# ... in which the girls total marks are one more than the boys

if sum(br) + 1 == sum(gr):

# all marks in top to bottom order

mks = sorted(br + gr, reverse=True)

# a girl is in seventh position

if mks[6] in gr:

f, s, l = ', '.join(str(m) for m in mks[:6]).rpartition(', ')

print(f"The top six marks are {' and '. join((f, l))}.")

Frits permalink

With the requirement of the lowest mark and an even average.


# return y // x if y is a divisor of x
divides = lambda x, y: [d for (d, r) in [divmod(x, y)] if not r]

# primes between 40 and 80
pb = set(n for n in range(41, 81, 2) if all(n % p for p in (3, 5, 7)))
        
# decompose <t> into 2 groups of arithmetic sequences, numbers from <ns>
def decompose(t, ns, k=6, s=[]):
  # do last 3 elements form an arithmetic progression?
  ar = lambda seq: 2 * seq[-2] == seq[-1] + seq[-3]
  
  if k == 1: # almost done
    if t > s[-1] and t in ns and ar(s_ := s + [t]):
      yield s_
  else:
    if k == 3:
      # check that first 3 numbers form an arithmetic progression 
      if not(ar(s)): return
    for i, n in enumerate(ns):
      if not s or n > (s[0] if k == 3 else s[-1]): 
        yield from decompose(t - n, ns - {n}, k - 1, s + [n])
        
# return first term for a specific constant ratio
# mx = int((100 / a)**.5 * a) + 1 = int(10 * a**.5) + 1 = 61
def first_term(cr, mn=37, mx=61):
  # geometric series
  gs = (1 + cr + cr**2) 
  # check totals of the form 12k + 1 (even average for boys)
  for t in range(6 * 36 + 12 + 1, int((36 + mx) * gs) + 1, 12):
    # t = (36 + d) * gs,  d = t / gs - 36
    d, r = divmod(t, gs)
    if r: continue
    d -= 36
    if not (mn <= d <= mx): continue
    yield int(d), t        

# check 2 groups (a, b, c) and (d, e, f)
a = 36  # lowest mark 
for b in range(a + 1, 61):   # 10 * a**.5 + 1
  for c in divides(b * b, a):
    # determine d of sequence (d, e, f) based on constant ratio b / a
    for d, t in first_term(b / a):
      for e in divides(b * d, a):
        for f in divides(b * e, a):   
          if len(girls := {a, b, c, d, e, f}) != 6: continue
          # collect 2 groups of arithmetic sequences for the boys
          for boys in decompose(t - 1, pb):
            # all marks in top to bottom order
            mks = sorted(set(boys) | girls, reverse=True)
            # a girl is in seventh position
            if mks[6] in girls:
              print(f"The top six marks are: {mks[:6]}")

# return y // x if y is a divisor of x

divides = lambda x, y: [d for (d, r) in [divmod(x, y)] if not r]

# primes between 40 and 80

pb = set(n for n in range(41, 81, 2) if all(n % p for p in (3, 5, 7)))

# decompose <t> into 2 groups of arithmetic sequences, numbers from <ns>

def decompose(t, ns, k=6, s=[]):

# do last 3 elements form an arithmetic progression?

ar = lambda seq: 2 * seq[-2] == seq[-1] + seq[-3]

if k == 1: # almost done

if t > s[-1] and t in ns and ar(s_ := s + [t]):

yield s_

else:

if k == 3:

# check that first 3 numbers form an arithmetic progression

if not(ar(s)): return

for i, n in enumerate(ns):

if not s or n > (s[0] if k == 3 else s[-1]):

yield from decompose(t - n, ns - {n}, k - 1, s + [n])

# return first term for a specific constant ratio

# mx = int((100 / a)**.5 * a) + 1 = int(10 * a**.5) + 1 = 61

def first_term(cr, mn=37, mx=61):

# geometric series

gs = (1 + cr + cr**2)

# check totals of the form 12k + 1 (even average for boys)

for t in range(6 * 36 + 12 + 1, int((36 + mx) * gs) + 1, 12):

# t = (36 + d) * gs, d = t / gs - 36

d, r = divmod(t, gs)

if r: continue

d -= 36

if not (mn <= d <= mx): continue

yield int(d), t

# check 2 groups (a, b, c) and (d, e, f)

a = 36 # lowest mark

for b in range(a + 1, 61): # 10 * a**.5 + 1

for c in divides(b * b, a):

# determine d of sequence (d, e, f) based on constant ratio b / a

for d, t in first_term(b / a):

for e in divides(b * d, a):

for f in divides(b * e, a):

if len(girls := {a, b, c, d, e, f}) != 6: continue

# collect 2 groups of arithmetic sequences for the boys

for boys in decompose(t - 1, pb):

# all marks in top to bottom order

mks = sorted(set(boys) | girls, reverse=True)

# a girl is in seventh position

if mks[6] in girls:

print(f"The top six marks are: {mks[:6]}")

Sunday Times Brain Teaser 1023 – A Mixed Maths Class

by Rachel Blunt

Published 7th March 1982 (see link)

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