# James Bond in trouble

Discussion in 'Mathematics' started by AVDBA, Sep 23, 2019.

1. ### AVDBANew commenter

James Bond is in trouble. He now has 7 extremely jealous girlfriends. 6 girlfriends are fine, but there is a problematic one that have decided to get rid of him. The agent thinks he now has fantastic girlfriends - when a message arrives. The message was jealously sent by secretary Miss Moneypenny:
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Dear Bond,

I know there is not my number within Your heart. From Monday to Sunday You have defined the fixed affairs of Yours... Going straight to the point, You are in danger, since 279999720 has a terrible plan against You. Sorry dear, but You have asked for training math, so I have defined 279999720 by adding up all of the numbers having k digits that are formed by the permutation without repetition using, as digits, the members of the set of natural numbers up to k. Your girlfriend k is Your danger and there is no place for nothing within my naturals.

Bye Bond,

Moneypenny
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Determine the day of his problematic girlfriend.

2. ### AVDBANew commenter

I have posted this problem by virtue of the simplicity one gains considering arguments on symmetry. It is a problem about permutations and requires the consideration of each element of permutation to be added, as defined by Moneypenny:

* ... so I have defined 279999720 by adding up all of the numbers having k digits that are formed by the permutation without repetition using, as digits, the members of the set of natural numbers up to k.

The members of the set of natural numbers up to k are these ones within this set: K = {0, 1, 2, ..., k} - {0} = {1, 2, 3, ..., k}, with k in {1, 2, 3, 4, 5, 6, 7} (by virtue of the days of the week, as explained beneath). Zero is being excluded by virtue of:

* ... there is no place for nothing within my naturals.

The quantity q of permutations without repetition is:

q = k!

Considering an arbitrary permutation, which is a number n having k digits:

n = ak ... a3 a2 a1

being a1, a2, ..., ak the digits of n - there exists another permutation, which is a number m having k digits:

m = [(k + 1) - ak] ... [(k + 1) - a3] [(k + 1) - a2] [(k + 1) - a1]

being [(k + 1) - a1], [(k + 1) - a2], [(k + 1) - a3], ..., [(k + 1) - ak] the digits of m. It is not difficult to infer that:

n + m = (k + 1)(10^0 + 10^1 + 10^2 + ... + 10^(k - 1))

implying that the number (n + m) has as its k digits the digit (k + 1), i.e., the natural number (k + 1) appears k times as a digit of the number (n + m):

n + m = [k + 1] [k + 1] ... [k + 1]

Hence, by adding up all of the numbers having k digits as defined by Moneypenny, one has:

(k + 1)(10^0 + 10^1 + 10^2 + ... + 10^(k - 1))(k!)/2 = 279999720

By factoring the number 279999720, one infers that, from the possibilities for the number (n + m) (777777, 66666, 5555, 444, 33, 2), that the possible factorization that fulfills the requirement of the problem is:

279999720 = 777777 x 360.

Hence:

k! = 2 x 360 = 720 => k = 6

* From Monday to Sunday You have defined the fixed affairs of Yours...

The day of Bond's problematic girlfriend is 6, i.e., Saturday, inferred from the above message sent by Moneypenny.

I hope You have enjoyed!