Mathematics Times – July 2019

3.Sol:

Miscellaneous

1.Sol:

2.Sol:

3.Sol:

1 2 3

x x x ... xn

n



   



and

2 2 2

x x 1 2 ... xn

n



  



Also given mean and standard deviations of

new list of numbers y 1 , y 2 ,.., yn are

ˆ y y^12 ... yn^0 x x^23 ... x x xn^11 n

n n

  

        

  

and

2 2 2

ˆ y y^12 ... yn

n



  



2 2 2 2 2

0 x x 2 3 ... xn 1 (x x 1 n)

n

      

2 2 2 2 2 2

0 x x 2 3 ... x x xn 1 1 n 2 x x 1 n

n

 

      

 

Therefore  ˆ.

3.Sol: Given information is represented in the
diagram as below

From the diagram, we get

1 3 3 5 2011 1

2 2 , 4 2 ,..., (^20122)
a a a a a a
a a a
  
  
That yields
2 2 2 ... 2a 2    a 4 a 6 a 2012   a a a 1 3 5
  ... a 2011 a 1
(^)      a a a 1 3 5 ... a 2011 a 1.
But we a a a 2      4 6 ... a 2012 3018.
Which yields that a a a 1      3 5 ... a 2011 3018.
Therefore the sum of all numbers is
a a a 1      2 3 ... a 2011 2(3018) 6036.
Miscellaneous
1.Sol: We know q 0 .Now we try to
prove
22
21
p
q


i.e
22
21 21
p
q 
we take
20 20
1 1
1 1 1
21 20 i i (^21) i i
p
q a
 a 
 
    
 

 

20 20

1 2 1

1 1 1

(^20) i 1 21i
i
i
 i  i
    
      

    

20 20

1 2 1

1 1 1

2 20 i 1 i 21

i

 i  i

  

     

   

20 20

1 2 1

1 1 1 1

2 20 2 i 1 i 21

i

 i  i

     

   

20 20

1 1

1 1 1 2 1

0 1 1

2 20 2 5 i (^21) i
     

  

1 1 1 2 1

19 20

2 20 2 5 21

       

 

 

1 1 22

1 8 1

2 20 21

    

2.Sol: Using Fermat‘s little theorem , we have,

If 5 |athen we have a^4  1 mod 5. That

is a^2   1 mod 5. If 5|z, then we are done

with z x y(^2 ^2 ). Otherwise we have

2 z^2   2 mod 5. That is    2 x y^22 , which

yields that x^2  1 mod 5 and y^2  1 mod 5 or

x^2   1 mod 5. Thus x y^2 ^2 is divisible by 5.

Likewise ,The same procedure can be done

with mod 3 of 3 | z. Otherwise if 3 divides z

then 3 divides x y^2 ^2 .Which yields that 3

divides x and y. so 3 divides x + y and x - y.

3.Sol: Here the phrase non-congruence triangle

represents counting un-ordered triplets. That

is the given condition is divided into three cases