Madam Mim has a deck of $52$ cards, stacked in a pile with their backs facing up. Mim separates the small pile consisting of the seven cards on the top of the deck, turns it upside down, and places it at the bottom of the deck. All cards are again in one pile, but not all of them face down; the seven cards at the bottom do, in fact, face up. Mim repeats this move until all cards have their backs facing up again. In total, how many moves did Mim make $?$
Problem
Source:
Tags: combinatorics, number theory
babu2001
06.05.2017 22:34
I think the problem asks for minimum number of moves, if so, I think the answer is $89\equiv -7^{-1}\pmod {52}$.
FedeX333X
07.05.2017 22:23
babu2001 wrote: I think the problem asks for minimum number of moves, if so, I think the answer is $89\equiv -7^{-1}\pmod {52}$. No, the right answer is 112.
khbghvb
08.12.2017 10:04
Are you sure about 112???
khbghvb
08.12.2017 10:05
Cause mine is more...
ironpanther29
17.08.2018 03:55
@above: Be nice!
FedeX333X
01.09.2018 17:04
khbghvb wrote: Are you sure about 112??? Yup, the official solution confirms that
tas
21.02.2020 10:55
The answer is 112
In 3434 ~ 3 (7 * 4, 8 * 3), 52 cards are divided into groups A and B. Each group A contains 4 cards and each group B contains 3 cards. Each operation flips a set of B and a set of A, and the relative positions of AB are ordered. It takes 14 turns to flip all A's and 16 turns to flip all B's. Total [14,16] = 112 times.