Teaser 3145: Easier to ask the audience
From The Sunday Times, 1st January 2023 [link] [link]
“I have forgotten the phone number!”, complained Martha, about to phone a friend. “So have I!”, replied George, “but I have some vague memories of it. It is a perfect square with all the digits different, and the last digit is equal to the number of digits to be dialled. The last-but-one digit is odd and one of the digits is zero. Also the second and third and last-but-one digits are all exact multiples of the first digit. Maybe you can work it out”.
Martha proceeded to dial the number correctly.
What number did she dial?
Happy New Year from S2T2!
[teaser3145]
GeoffR, 
Frits, and 1 other are discussing. 
Jim Olson and 
Hugh Casement, 
Tony Brooke-Taylor, 
Dirk, 
Tony Brooke-Taylor, 
S2T2 
Jim Randell 4:47 pm on 30 December 2022 Permalink |
The final digit of the square number tells us how long it is, so we only need to check up to 9 digits, and it must have a “second”, “third”, and “last-but-one” digit, so (assuming these are all different) there need to be at least 5 digits. And to also fit a 0 digit in there must be at least 6 digits.
And a perfect square cannot end with 7 or 8, so we only need to consider 6 or 9 digit numbers.
This Python program just tries all candidate squares in this range (and we could add further analysis to reduce the number of candidates).
It runs in 122ms. (Internal run time is 23.3ms).
Run: [ @replit ]
from enigma import (powers, inf, nsplit, printf) # is <x> a (proper) multiple of <n>? is_multiple = lambda n, x: (x > n > 0 and x % n == 0) # consider squares with 6 - 9 digits for n in powers(317, inf, 2): # split into digits ds = nsplit(n) k = len(ds) # there are at most 9 digits... if k > 9: break # but not 7 or 8 if not (k == 6 or k == 9): continue # ... and they are all different if not (len(set(ds)) == k): continue # the final digit is equal to the total number of digits if not (ds[-1] == k): continue # the last but one digit is odd if not (ds[-2] % 2 == 1): continue # and the digit 0 is used if not (0 in ds): continue # the 2nd, 3rd and last but 1 digit are multiples of the 1st digit if not all(is_multiple(ds[0], ds[i]) for i in (1, 2, -2)): continue # output solution printf("n = {n}")Solution: The number is: 173056.
173056 = 416².
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Jim Randell 10:37 pm on 31 December 2022 Permalink |
With a bit of analysis we see that a 6-digit square must be of the form 1xx0x6, and so is the square of a number in the range [351, 445] that ends in the digit 4 or 6.
And a 9-digit square must be of the form 1xxxxxxx9, and so is the square of a number in the range [11093, 13698] that ends in the digit 3 or 7. (Thanks to Frits for correcting my upper bound).
This Python program just checks these numbers.
It runs in 52ms. (Internal runtime is 1.2ms).
Run: [ @replit ]
from enigma import (irange, nsplit, printf) # is <x> a (proper) multiple of <n>? is_multiple = lambda n, x: (x > n > 0 and x % n == 0) def check_n(n): # split into digits ds = nsplit(n) k = len(ds) # all digits are all different if not (len(set(ds)) == k): return # the final digit is equal to the total number of digits if not (ds[-1] == k): return # the last but one digit is odd if not (ds[-2] % 2 == 1): return # and the digit 0 is used if not (0 in ds): return # the 2nd, 3rd and last but 1 digit are multiples of the 1st digit if not all(is_multiple(ds[0], ds[i]) for i in (1, 2, -2)): return # output solution printf("n = {n}") def check(seq): for i in seq: check_n(i * i) # check 6-digit numbers of the form 1xx0x6 check(irange(354, 444, step=10)) check(irange(356, 436, step=10)) # check 9-digit numbers of the form 1xxxxxxx9 check(irange(11093, 13693, step=10)) check(irange(11097, 13697, step=10))LikeLike
Frits 11:07 am on 1 January 2023 Permalink |
@Jim, I came to the same conclusion but with ranges [351-445] (you probably had a typo) and [11093-13698].
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Jim Randell 11:16 am on 1 January 2023 Permalink |
Thanks. Yes, the upper bound should be 13698.
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Tony Smith 2:17 pm on 1 January 2023 Permalink |
The only squares with odd penultimate digits have last digit 6.
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Jim Randell 2:21 pm on 1 January 2023 Permalink |
Neat. So we only need to check 9 cases:
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Frits 3:25 pm on 1 January 2023 Permalink |
@Jim, the check(irange(354, 444), step=10) is still needed.
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Jim Randell 9:19 pm on 1 January 2023 Permalink |
Right. Still, it is only 19 cases to check. Easy enough to do on a pocket calculator.
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Jim Randell 2:10 pm on 2 January 2023 Permalink |
Here is a version that uses less brute force (and less analysis).
The final two digits of the square number depend only on the final two digits of its root, so we can consider possible values for these 2 digits, and find those that give acceptable final two digits of the square. The final digit must corresponding to the digit length of the square, and then penultimate digit must be odd.
We then know the length of the square, final two digits and the leading digit must be a divisor of the penultimate digit, so we can consider possible roots that give the required conditions.
In the end we only need to apply the remaining tests on a handful of values.
This Python program runs in 51ms. (Internal runtime is 368µs).
Run: [ @replit ]
from enigma import (irange, inf, sq, nsplit, sqrt, ceil, printf) # is <x> a (proper) multiple of <n>? is_multiple = lambda n, x: (x > n > 0 and x % n == 0) # consider final 2 digits of the root of the perfect square for i in irange(0, 99): # and calculate the final 2 digits of the square (y, z) = nsplit(sq(i), 2) # y must be odd if y % 2 != 1: continue # z counts the total number of digits, which cannot be less than 5 if z < 5: continue # y is a multiple of the leading digit (a) for a in irange(1, y // 2): if y % a != 0: continue # consider possible roots lo = ceil(sqrt(a * pow(10, z - 1)) - i, 100) for j in irange(lo, inf, step=100): n = sq(i + j) ds = nsplit(n) # there must be z digits if len(ds) < z: continue if len(ds) > z: break # and it must start with a if ds[0] < a: continue if ds[0] > a: break # there must be a 0 digit if 0 not in ds: continue # each digit must be different if len(set(ds)) != z: continue # 2nd and 3rd digits must be multiples of the 1st if not (is_multiple(a, ds[1]) and is_multiple(a, ds[2])): continue # output solution printf("n = {n} [{r}^2]", r=i + j)LikeLiked by 1 person
GeoffR 10:25 am on 31 December 2022 Permalink |
# check 5 - 9 digit square numbers sq = [x * x for x in range(100, 31622)] for n in sq: # check the number of digits are all different if len(str(n)) == len(set(str(n))): n_str = str(n) # check last-but-one digit is odd # ... and one of the digits is zero after the third digit if int(n_str[-2]) % 2 == 1 and '0' in n_str[2:]: # check 2nd and third digits are a multiple of the 1st digit if int(n_str[1]) % int(n_str[0]) == 0: if int(n_str[2]) % int(n_str[0]) == 0: # check last-but-one digit is a multiple of the 1st digit if int(n_str[-2]) % int(n_str[0]) == 0: # check last digit is length of number dialled if int(n_str[-1]) == len(str(n)): print(f"Number dialled was {n}.")LikeLike
GeoffR 11:32 am on 2 January 2023 Permalink |
This solution uses previous postings in that the telephone number is six or nine digits long and the possible ranges of square values.
from enigma import nsplit, all_different # check six digit phone numbers (abcdef) n = 353 while n < 445: n += 1 n1 = n * n # potential square tel no. a, b, c, d, e, f = nsplit(n1) # teaser digit conditions if f == 6 and d == 0 and e % 2 == 1: if all_different(a, b, c, d, e, f): # 2nd, 3rd and penultimate digits #...are multiples of 1st digit if all(x % a == 0 for x in (b, c, e)): print(f"Martha dialled {n1}.") break # check nine digit phone numbers (abcdefghi) m = 11092 while m < 13699: m += 1 m1 = m * m # potential square tel no. a, b, c, d, e, f, g, h, i = nsplit(m1) # teaser digit conditions if i == 9 and 0 in (d, e, f, g) and h % 2 == 1: if all_different(a, b, c, d, e, f, g, h, i): # 2nd, 3rd and penultimate digits # ...are multiples of 1st digit if all(x % a == 0 for x in (b, c, h)): print(f"Martha dialled {m1}.") breakLikeLike
GeoffR 5:20 pm on 2 January 2023 Permalink |
This program checks for a 6-digit solution only.
No 9-digit solution was found in previous programs.
from enigma import is_square D, F = 0, 6 # values fixed by teaser conditions for A in range(1, 10): for B in range(1, 10): if B == A:continue if B % A != 0:continue for C in range(1, 10): if C in (A, B):continue if C % A != 0:continue for E in range(1, 10): if E in (A,B,C,F):continue if E % 2 != 1 or E % A != 0: continue num = A * 10**5 + B * 10**4 + C * 10**3 + E*10 + F if num // 100 % 10 != D:continue if not is_square(num):continue print(f"Martha dialled {num}.")LikeLike