From The Sunday Times, 26th April 1970 [link]
Near my house is an impressive block of flats. It stands 16 storeys high, and each story is a lofty 15 feet from floor to floor. My old friends Bob and Horace operate the two lifts, and both start their shift each morning leaving the ground floor at 8 a.m. precisely.
The two lifts make automatic compulsory stops at the ground level (there is no basement), the 12th and top floors. But below the 12th, Bob serves only the odd-numbered floors, and Horace the even numbers; these stops are also compulsory (up and down). All stops take just 10 seconds, except at ground level where both lifts wait for 20 seconds.
Above the 12th both Bob and Horace stop as desired, going up and coming down. Both lifts travel between floors at identical speeds.
Every morning the two attendants make a rendezvous to exchange newspapers and collect their coffee when their arrivals at a certain floor coincide exactly at two minutes after 11 a.m.
In feet per second what the speed of each lift between stops?
Note: In the UK the floors in buildings are: ground floor, first floor, second floor, etc.
Also, as stated this puzzle does not appear to have a unique solution.
An answer was published (with Teaser 467) stating:
“This clever trap, with its nice reasoning, outweighed the slight mathematical bias and kept down the entry to a mixed handful.”
But later, with Teaser 471 the following statement was made:
“[465: Regretfully a wrong transposition led to a false calculation of time.]”
In the comments I give a variation on the puzzle that does have a unique answer.
This puzzle was originally published with no title.
[teaser465]
Frits, 
GeoffR, 
Lise Andreasen and 
S2T2 
Jim Randell 9:26 am on 5 April 2019 Permalink |
The product has to be long enough to accommodate two pairs of different digits in, so it must be at least 4 digits long.
But the total multiplication sum only uses 10 digits, so the three original numbers can use at most 6 digits, and as the number of digits in each multiplicand is different they must use at least 6 digits, so the sum looks like one of these:
We can solve these using the [[
SubstitutedExpression()]] solver from the enigma.py library (although it is not especially quick). The following run file executes in 619ms.Run: [ @replit ]
Solution: The product is 8778.
The full sum is:
The solution uses the
XYYXpattern for the product, and is the only solution using that pattern.However, without the restriction that A is greater than 1 we can find a further solution using the
XXYYpattern:And there are no solutions that use the remaining
XYXYpattern. (This can be seen by observing that the product has a prime factor of 101, and we cannot have a three digit multiple of 101 that does not have repeated digits).LikeLike
Jim Randell 10:35 am on 5 April 2019 Permalink |
Here is a faster solution in Python. It considers possible values for the result of the multiplication, and then factors the result into 1-,2-,3-digit numbers.
It runs in 96ms.
from enigma import (irange, subsets, divisors_pairs, nsplit, printf) # choose X and Y for (X, Y) in subsets(irange(0, 9), size=2, select='P'): if X == 0: continue # consider possible results for (i, j) in [(1100, 11), (1010, 101), (1001, 110)]: n = X * i + Y * j # find 1-, 2-, 3- digit factors of n for (A, m) in divisors_pairs(n): if A < 2: continue if A > 9: break if len(set((A, X, Y))) != 3: continue for (BC, DEF) in divisors_pairs(m): if BC < 10: continue if BC > 99: break if not (99 < DEF < 1000): continue (B, C) = nsplit(BC) (D, E, F) = nsplit(DEF) if len(set((A, B, C, D, E, F, X, Y))) != 8: continue # output solution printf("{A} * {BC} * {DEF} = {n}")LikeLike
GeoffR 11:28 am on 11 April 2019 Permalink |
LikeLike
GeoffR 8:58 am on 8 February 2024 Permalink |
# Sum digit pattern is A × BC × DEF = XXYY, XYXY or XYYX # LHS of sum for A in range(2, 10): for BC in range(12, 99): B, C = BC //10, BC % 10 if A in (B, C):continue for DEF in range(123, 988): D, E, F = DEF // 100, DEF // 10 % 10, DEF % 10 if len({A, B, C, D, E, F}) == 6: # RHS of sum prod = A * BC * DEF if not 1000 < prod < 9999:continue M, N = prod // 1000, prod // 100 % 10 P, Q = prod // 10 % 10, prod % 10 # Digit pattern of MNPQ is XXYY, XYXY or XTTX if len({M, N, P, Q}) == 2: if (M == N and P == Q) or (M == P and N == Q) \ or (M == Q and N == P): if len({A, B, C, D, E, F, M, N, P, Q}) == 8: print(f"Sum: {A} x {BC} x {DEF} = {prod}.") # Sum: 3 x 14 x 209 = 8778.LikeLike
Frits 3:21 pm on 8 February 2024 Permalink |
# dictionary of product a * bc with different digits (not ending on 1) d = dict() for i in range(2, 10): for j in range(10, 50): if i < j and i * j < 100 and (i * j) % 10 != 1 and j % 10 and \ len(set(ij := str(i) + str(j))) == len(ij): d[i * j] = d.get(i * j, []) + [(str(i), str(j))] # A * BC * DEF = RHS # RHS = XYXY is not possible as XYXY is a multiple of 101 # A number is divisible by 11 if the difference between the sum of its digits # in odd places and the sum of the digits in even places is either 0 or # a multiple of 11 # thus RHS is a multiple of 11 for XXYY and XYYX for i in range(10, 91): DEF = 11 * i sDEF = str(DEF) if sDEF[-1] == '0' or len(set(sDEF)) != 3: continue # ABC * DEF = RHS for ABC in range(max(20, (999 // DEF) + 1), (9999 // DEF) + 1): sRHS = str(ABC * DEF) if len(set(sRHS)) != 2 or any(x in sRHS for x in sDEF) or \ sRHS.count(sRHS[0]) != 2: continue # is ABC a valid product if ABC not in d: continue for a, b in d[ABC]: # different digits if len(set(s := (a + b + sDEF + sRHS))) != len(s) - 2: continue print(f"answer: {sRHS} ({a} * {b} * {DEF})")LikeLike