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A student from Mearns Castle made the
useful connection with another problem published this
This is similar to finding the number of lines in Mystic Rose
Say there are n people. The first person shakes his hand with the
other n-1 people.
The second person then shakes his hand with the other n-2
And so on until the (n-1)th person shakes his hand with the nth
So the number of handshakes is (n-1) + (n-2)... + 3 + 2 + 1 which
Another way of considering it is that each person has a total of
n-1 handshakes, and there are n people, so there would be (n-1)(n)
handshakes, but this includes each handshake twice (1 and 2, 2 and
1) so dividing by 2 gives the correct answer.
Aswaath from the Garden International
School in Kuala Lumpur, Malaysia mentioned that the method for
solving this problem connected with the
method used by Gauss when he was still
a young student.
Joe from Hove Park Lower School also
noticed connections with other work:
If, for example, 10 mathematicians met, the first will make 9
handshakes, the second makes 8, the third makes 7 and so on until
the tenth finds he has already made handshakes with everyone and so
makes no more.
This gives 9+8+7+6+5+4+3+2+1+0 handshakes and this is 45.
But look at the sequence... it is the 9th triangular
Triangle Numbers and/or Clever Carl)
The formula for the Tth triangular number is T(T+1)/2
With the handshake problem, if there are n people, then the number
of handshakes is equivalent to the (n-1)th triangular number.
Subsituting T = n-1 in the formula for triangular numbers, we can
deduce a formula for the number of handshakes between n
Number of handshakes = (n-1)(n)/2
Jayme from the Garden International
School agreed and used this insight to correct Sam's
Sam's method isn't right because he did not divide the answer by 2.
If you do not divide it by 2, you will be counting the number of
handshakes between pairs of mathematicians twice.
If 20 mathematicians meet:
20 x 19 = 380
380 / 2 = 190 (total number of handshakes)
If 161 mathematicians meet:
161 x 160 = 25760
25760 / 2 = 12880 (total number of handshakes)
Could there be exactly 4851 handshakes at a gathering where
everyone shakes hands?
We know that 4851 handshakes is between 20 mathematicians and 160
mathematicians. So, let's start by trying out 100
Total number of handshakes = 100 x 99 / 2 = 4950
That's too many so now I'll try 99 mathematicians.
Total number of handshakes = 99 x 98 / 2 = 4851
Therefore, there could be exactly 4851 handshakes when 99
Could there be exactly 6214 handshakes at a gathering where
everyone shakes hands?
Let's start by trying out 112 mathematicians.
Total number of handshakes = 112 x 111/ 2 = 6216
That's too many so now I'll try 111 mathematicians.
Total number of handshakes = 111 x 110 / 2 = 6105
That's too few so there cannot be exactly 6214 handshakes.
Joseph from Bradon Forest School and
Tabitha from The Norwood School used similar reasoning:
Sam's reasoning is wrong because you should only count the unique
handshakes that have taken place. He should modify his method by
dividing by 2. Each handshake involves 2 mathematicians so only
half of Sam's handshakes are unique.
Total handshakes = $161 \times 160/2 = 12880$
$N$ = number of mathematicians
Total handshakes = $N\times (N - 1)/2$ or $(N^2 -N)/2$
Can $4851$ be the exact number of handshakes at a gathering?
$(N^2 -N)/2 = 4851$
$(N^2 -N)= 9702$
$9702 = approx. 10000 = 100^2$
Try $N = 99$
$99^2 -99 = 97024$
So yes, if $99$ mathematicians meet there will be $4851$
No, the nearest triangle number is $6216$, and $6214$ is not a
Yes, $86$ Mathematicians
Yes, $124$ Mathematicians
No, the nearest triangle number is $8646$, and $8656$ is not a
We received many more correct solutions,
including very clear ones from Siddhartha and Tasuku
from the Garden International School in Kuala
Lumpur, Ben from Bedminster Down School,
Abhinav from Bangkok Patana School and Luke from Maidstone Grammar
School. Well done and thank you to you all.