Sunday Times Teaser 3145 – Easier To Ask The Audience
by Danny Roth
Published Sunday January 01 2023 (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?
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Brian Gladman permalink1234567891011121314151617181920212223from math import isqrt# the set of unit digits of squares with odd penultimate digitsunits = set(i ** 2 % 10 for i in range(10) if (i ** 2 // 10) & 1)# consider the possible lengths of the phone numberfor sq_len in units:# digits at positions 0, 1, 2, -2, -1 all non zero and different# plus one zero digit gives a minimum of six digitsfor r in range(317, isqrt(10 ** sq_len - 1) + 1):nbr = r ** 2# its penultimat digit is oddif (nbr // 10) & 1:# the digits of the numberd = [int(d) for d in str(nbr)]# its digits are all different with one zero ddigitif len(set(d)) == sq_len and d.count(0) == 1:# its second, third and last but two digits are exact# proper multiples of its first digitif all(n > d[0] and n % d[0] == 0 for n in (d[1], d[2], d[-2])):print(f"She dialled {nbr}.")
