Teaser 2565: Red and black
From The Sunday Times, 20th November 2011
I have taken 10 cards and written a different digit on each one. Some of the cards are red and the rest are black. I have placed some of the red cards in a row to form a long number. Then I have moved the last card to the front to give me a larger number. In fact this larger number divided by the original one equals a digit on one of the black cards.
What was the original number?
[teaser2565]
Jim Randell 3:53 pm on 11 February 2019 Permalink |
This puzzle is similar Enigma 1036.
Here is a Python program that solves it using a similar analysis.
Run: [ @repl.it ]
from enigma import irange, nsplit, int2base, printf # consider multipliers for k in irange(2, 9): q = 10 * k - 1 # consider n digit numbers for n in irange(1, 8): p = 10 ** n - k # consider possible mobile digit d (different from k) for d in irange(0, 9): if d == k: continue # the corresponding n digit number x, is... (x, r) = divmod(d * p, q) if r != 0: continue # x has to be n different digits s = set(nsplit(x, n)) if len(s) != n: continue # also different from d and k if d in s or k in s: continue printf("{x}{d} x {k} = {d}{x}", x=int2base(x, width=n))Solution: The original number is 230769.
So we have:
If we allow leading zeros there is a further solution:
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