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 Ady's RH problem (Posted on 2014-03-11)
Let

Sequence1= (all existing 4-digit numbers, no leading zeros);
Sequence2= (each member of seq1 replaced by 1 iff fulfilling both conditions : cond1 and cond2; replaced by 0 otherwise.)

Cond1: k (the number of letters in the English set of words representing N) is even.
For example the number 3024 (three thousand twenty four) does not qualify (23 letters) .

Cond2: the number N is evenly divisible by 3, e.g. the number 3024 qualifies.

Ergo: 3024 of Sequence1 will be replaced by 0 in Sequence2.

Denote by N1 the number of ones in the seq2 and by N0 the number of zeroes in that sequence.

Don't be intimidated by the D4 label,- reread the text carefully and try to answer all of the following questions:

a. What does the RH stand for?
b. Is it true that abs(N1- N0) is an even number?
c. What is the meaning of KISS?
d. Why D4?

 See The Solution Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Half-serious answer | Comment 1 of 6

The only loose connection I can make is that despite laying out the criteria for determining the sequences, at no point did you really ask anything that requires actually determining the sequences.  (Sadly I only realized this after working out sequence 2, anyway).

So despite appearing "really hard," at the outset, if you "keep it simple", read carefully and just answer the questions being asked, it's actually quite easy.

a) "Really Hard"

b) Yes - regardless of how you determine Sequence 2, N1 is simply 9000 - N0.  So abs(N1 - N0) = abs(9000 - N0 - N0) = abs(9000 - 2N0) which is clearly even regardless of what N0 is.

c) "Keep it simple, stupid."

d) See (a).

:)

 Posted by tomarken on 2014-03-11 15:25:35

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