Solution to Mark Zuckerberg's problem

FOR SEE THE QUESTION CLICK HERE

You cannot take 9 different value that's difference cannot be divisible by 8,

suppose i have 9 different age a,b,c,d,e,f,g,h,i and whose difference is

x1=(a-b)   y1=(b-c)  z1=(c-d)  p1=(d-e)  q1=(e-f)   r1=(f-g)  s1=(g-h)  t1=(h-i)
x2=(a-c)   y2=(b-d)  z2=(c-e)  p2=(d-f)   q2=(e-g)  r2=(f-h)  s2=(g-i)
.                .                 .                .                  .        r3=(f-i)                    
.                .                 .                .                  .                                              
.                .                 .                .                  .                                            
x8=(a-i)   y7=(b-i)   z6=(c-i)  p5=(d-i)  q4=(e-i)  

now let's see how many pair is possible,
for 'a' there is 8 as shown above,
for 'b' there is 7,because '(b-a)' is repeated
for 'c' there is 6,
.
.
for 'h' there is only one pair i.e. (h-i)
so 8+7+6+5+4+3+2+1=36 pair possible


now take modulo operation of 8 on x1 to x8,y1 to y7,z1 to z6,p1 to p5,q1 to q4,r1 to r3,s1 to s2 and t1
you will get all value between 0 to 7

now let's understand how we get value between 0 to 7,
for,
[0]= value should be {0,8,16,24,...}

[1]= value should be {1,9,17,25,...}

[2]= value should be {2,10,18,26,...}

[3]= value should be {3,11,19,27,...}

[4]= value should be {4,12,20,28,...}

[5]= value should be {5,13,21,29,...}

[6]= value should be {6,14,22,30,...}

[7]= value should be {7,15,23,31,...}

now if we get "0" in our x1 to x8 differences than it is divisible by 8,and if we do not get "0" anywhere then there must be some repeated number b/w 1 to 7 in x1 to x8.
so difference b/w that number is "0" and that is divisible by 8.

let's take on e.g.
Suppose,
x1=1
x2=2
x3=3
x4=4
x5=6
x6=5
x7=7
x8=4(repeated)

now difference b/w x8 and x4 is "0",so x8 - x4=(a-i) -(a-d)=(d-i)=p5 

So we got one pair that is divisible by 8.

Have a nice day....

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