Sunday Times Teaser 2899 – Base Jumping

by Stephen Hogg

Published April 15 2018 (link)

In written multiplication tests Jay always calculated correct answers, but, quirkily, wrote the digits of any two-figure answers in reverse order. A completed test sheet (pre-printed with ordered triplets of single-figure positive integers to multiply, eg, 1x2x2= and 3x5x9=) included a question and his two-figure written answer that would be right when thought of in a number base other than ‘base 10’ (any numerals keeping their usual values). Jay told Kay about this, but didn’t give the question or answer. Kay asked whether one of that question’s numbers to multiply was a particular value and from the reply knew both the question and the answer Jay had written.

What was Jay’s written answer to that question?

2 Comments Leave one →

Brian Gladman permalink


from itertools import combinations_with_replacement
from collections import defaultdict

# index solutions on each of their multipliers
n2s = defaultdict(list)
# consider all combinations of three (possibly repeated)
# ordered single digits
for i, j, k in combinations_with_replacement(range(1, 10), 3):
  # find the tens and units digits in decimal notation
  t, u = divmod(i * j * k, 10)
  # consider only two digit results (in either base)
  if 0 < t < 10 and 0 < u < 10:
    # now find a base in which reversing the decimal digits
    # gives a correct answer (i.e. u * b + t =  10 * t + u)
    b, r = divmod(9 * t + u, u)
    # the digits must be less than the base 
    if not r and max(t, u) < b:
      # save the result indexed on each of the three multipliers
      tu = ((i, j, k), t, u, b)
      n2s[i].append(tu)
      n2s[j].append(tu)
      n2s[k].append(tu)

# now look for a multiplier for which the result is unique    
for key in n2s.keys():
  ((i, j, k), t, u, b), *r = n2s[key]
  if not r:
    print(f'{u}{t} ({i} x {j} x {k} = {t}{u} base 10 = {u}{t} base {b}).')

from itertools import combinations_with_replacement

from collections import defaultdict

# index solutions on each of their multipliers

n2s = defaultdict(list)

# consider all combinations of three (possibly repeated)

# ordered single digits

for i, j, k in combinations_with_replacement(range(1, 10), 3):

# find the tens and units digits in decimal notation

t, u = divmod(i * j * k, 10)

# consider only two digit results (in either base)

if 0 < t < 10 and 0 < u < 10:

# now find a base in which reversing the decimal digits

# gives a correct answer (i.e. u * b + t = 10 * t + u)

b, r = divmod(9 * t + u, u)

# the digits must be less than the base

if not r and max(t, u) < b:

# save the result indexed on each of the three multipliers

tu = ((i, j, k), t, u, b)

n2s[i].append(tu)

n2s[j].append(tu)

n2s[k].append(tu)

# now look for a multiplier for which the result is unique

for key in n2s.keys():

((i, j, k), t, u, b), *r = n2s[key]

if not r:

print(f'{u}{t} ({i} x {j} x {k} = {t}{u} base 10 = {u}{t} base {b}).')

Jim Randell permalink

My code is similar to Brian’s, but I think we need to consider if Kay received a “No” answer, so I record the potential solutions by each divisor and a Yes/No answer before looking for a unique key.


from collections import defaultdict

# accumulate results by (<divisor>, <answer>) pairs
s = defaultdict(list)

# consider triples composed of single digits (i, j, k)
for i in range(1, 10):
  for j in range(i, 10):
    for k in range(j, 10):
      n = i * j * k
      if n > 99: break
      (x, y) = divmod(n, 10)
      if x == y or y == 0: continue

      # n = 10x + y = by + x
      (b, r) = divmod(n - x, y)
      if r != 0 or not(x < b and y < b): continue

      # record tuples
      t = (i, j, k)
      print(f"[{n} (base 10) = {y}{x} (base {b}), {n} -> {t}]")
      for d in range(1, 10):
        s[(d, d in t)].append((n, t, x, y, b))

# look for keys (divisors) that give unique sums
for (k, v) in s.items():
  if len(v) > 1: continue
  (n, t, x, y, b) = v[0]
  print(f"question: {t[0]}x{t[1]}x{t[2]} (= {n}), jay's answer: {y}{x} (in base {b} = {n}) [k={k}]")

from collections import defaultdict

# accumulate results by (<divisor>, <answer>) pairs

s = defaultdict(list)

# consider triples composed of single digits (i, j, k)

for i in range(1, 10):

for j in range(i, 10):

for k in range(j, 10):

n = i * j * k

if n > 99: break

(x, y) = divmod(n, 10)

if x == y or y == 0: continue

# n = 10x + y = by + x

(b, r) = divmod(n - x, y)

if r != 0 or not(x < b and y < b): continue

# record tuples

t = (i, j, k)

print(f"[{n} (base 10) = {y}{x} (base {b}), {n} -> {t}]")

for d in range(1, 10):

s[(d, d in t)].append((n, t, x, y, b))

# look for keys (divisors) that give unique sums

for (k, v) in s.items():

if len(v) > 1: continue

(n, t, x, y, b) = v[0]

print(f"question: {t[0]}x{t[1]}x{t[2]} (= {n}), jay's answer: {y}{x} (in base {b} = {n}) [k={k}]")

In the end there is only one possible solution, where Kay asks about the digit 4, and receives a “Yes” answer.

I found it curious that we are asked to submit Jay’s written answer, which can refer to multiple possible written questions, rather than the three digits of the written question, which are unique (and from which we can determine Jay’s answer).

Sunday Times Teaser 2899 – Base Jumping

by Stephen Hogg

Published April 15 2018 (link)

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