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I confirm Elainavw's values. coolbee's value can not be correct since both of us used integer mathematics and the denominator can not vanish into thin air.
What is going on here?Can any sane person explain to me?
Hi;
I found another answer, but still for X = 4.
It uses 12 less "4"s than in my previous solution.
I tried for X = 3 again, but can't find a solution for it.
Now it is more picturesque than before.
In this I implemented value 3 from the top right to bottom left diagonal. Here also totals for P and Q are same value being 78. I had thought in the previous case it was because of actual crossing of diagonals. But it is not so.So it becomes necessary to go for 4 handshakes. it means minimum number of X asked in the puzzle is 4.
Thanks Anna, correction made.I was doing routine copying from one cell.Hence the mistake.
When we have finished with "P" and are going to "Q" we find another 3 can not be added as the diagonal joining (f,B) to (B,f) falls into a ditch ("x" of the main diagonal )and the total for Q remains as that of "P" viz. 77. this difficulty may be overcome by revising Q's handshake with "f" to 4 instead of 3.That should be continued to all the persons from "R" to "e".
It seems i put some junk value in the numerator of the fraction by mistake.It can not end with 5.
Last night I had an idea of what happened. We started with A having 2 handshakes with all.with b all but the last "f" had 2 handshakes and with "f" 3 handshakes. So the wave of "3" advanced leftwards on a lesser than longest backward diagonal. in the chess board terminology this diagonal had the same color as the main right diagonal containing "x" value. So the two diagonals crossed and the 16th man(Q)and 17th man(R) had to have same no. of handshakes. If we had started with 3 on the right top corner the two longest diagonals do not intersect and the problem does not arise. I was to post the same today morning but you were one step ahead of me.We can repair the previous set up by using 4 handshakes with "f" for everybody from "Q" to "e".
In your first line under "f" the term should be 2;Please see the symmetrical 1st point on the last line for "f" under "A" It is 2. The sheet is symmetrical albout the main diagonal.The proposal is "A" will have 2 handshakes with all others including "f".
A will have 2+2+2 ... 31 times=62
B will have 30*2+3=63
C will have 29*2+2*3=64
"e" will have 2*2+29*3=91
"f" will have 2+30*3=92
Why phrontister's excel table does not have row for A ? and why in the main diagonal "x" is not put?
I change my answer. minimum X=3
In phrontister's excel table in the first line (corresponding to B) under f 3 is o.k.
In the 2nd line we will keep 3 under "e" and 3 under "f".
In the 3rd line 3 under "d","e' and "f".
"f" will not have heavy burden. the totals under A to e will remain same. Only total of "f" will decrease to 2+30*3=92
I am wondering why it is not 93.i have to check the table.
I see. I misunderstood the conditions.Now I have to start afresh.
The irony is those who are having energy refuse to work.
32 people were invited at a party and started exchanging handshakes. Because of the confusion, each of them shook hands with each other multiple times: at least twice and up to X times. However, every invitee exchanged different number of handshakes from every other invitee. What is the minimum possible number X, so that the above condition is met?
As per the terms of problem setter, If A has 10 handshakes with P, he can not have 10 handshakes with anybody else. At the same time P also had 10 handshakes with A ;so P also can not have 10 handshakes with anybody else. here 10 handshakes means 10 and 10 only.Neither more nor less.
It is not clear whose brain is scanned and what is the result? What is the probability that the prediction can go wrong?Is the predictor independent or is associated with manipulating the contents of the box?
what I assumed is that a must have different number of handshakes from others but these numbers could be used by others also. e.g. if A had 10 handshakes with P ,R can have 10 handshakes with C ;there is no bar for that but obviously P can not have 10 handshakes with anybody other than A.
If there is a single variable the powers must be different.
Since the minimum returns is $2 it is worth $2. Anything you get above it is your profit.
A had 2 handshakes with B,3 handshakes with C, 4 handshakes with D .... 32 handshakes with f (32nd person).So total handshakes by A=2+3+3+4+.......+31+32=527 handshakes. Add handshakes of all people and divide by 2.
Is it a fiction?
My calculation agrees with bobbym. I calculated the sum on excel sheet.My value is 226494. coolbee's value comes to 191193.3048. coolbee, please check the formula.
The total number of handshakes is 8672.