online user counteronline user counter

MATHMAA

Only Search Your Site

16th KVJ Solutions

Probability Introduction :

16th KVJ Solutions contains Q & A for all in detail . Please feel free to ask any question or to give suggestion .

1Q) Prove that for every prime P>7, \(P^{6}-1\) is divisible by 504.

Solution : 504 = \(2^{3}.3^{2}.7\)

  To prove that \( P^{6}-1\) is divisible by \( 2^{3}.3^{2}.7\).

 i.e ., it is enough to show that \( P^{6}-1\) is a multiple of \(2^{3}=8\), \(3^{2}=9\) and 7 as there is no common factor for 8, 9, and 7 other than 1.

Case i :

  To prove that \(P^{6}-1\) is multiple of 8.

\( P^{6}-1=(P^{2})^{3}-1 \)

                =\( (P^{2}-1)(P^{4}+P^{2}+1) \)

                =\( (P+1)(P-1)(P^{4}+P^{2}+1)\)

P > 7 and P is a prime number.

P-1, P+1 are even numbers and one is multiple of 2 and other is 4.

   a) Let P-1 =4k+2  then P+1 = P-1+2

                                      =4k+2+2

                                       =4k+4, it is multiple of 4.

If P-1 is a multiple of 2 then P +1 is multiple of 4.

  b) Let P - 1 is 4 multiple , to show that P+1 is 2 multiple.

    Let P- 1 = 4k

         P + 1 =P-1 +2

                  =4k+2 , it is 2 multiple.

 Therefore if P-1 is 4 multiple then P +1 is 2 multiple.

(P-1)(P+1) is a multiple of 4*2=8

Hence \(P^{6}-1\) is a multiple of 8.

Case ii:

 To show that \( P^{6}-1\) is a multiple of 9.

We know that product of 3 consecutive numbers is divisible by 6 .

Product of 3 consecutive numbers is also divisible by 3.

P-1, P, P+1 are 3 consecutive numbers.

(P-1)(P)(P+1) is divisible by 3.

P is not divisible by 3 as P >7 and prime number.

So, (P-1)(P+1) is divisible by 3

\(P^{2}-1\) is a multiple of 3.

\(p^{6}-1=(p^{2}-1)(P^{4}+P^{2}+1) \)

                =\((P^{2}-1)((P^{2}-1)^{2} +3P^{2})\)

  \( (P^{2}-1)^2 +3P^{2}\) is a multiple of 3

Hence \((P^{2}-1)((P^{2}-1)^{2} +3P^{2})\) is a multiple of 3*3=9

Therefore\( P^{6}-1\) is a multiple of 9.

Case iii :

To prove that \( P^{6} -1\) is a multiple of 7.

P is prime so using Fermats theorem we get 

\( P^{6} = 1 mod 7\)

means \(P^{6}-1\) is a multiple of 7 .

Hence \(P^{6}-1\) is a multiple of 8*9*7=504 .

Therefore \(P^{6}-1\) is divisible by 504.

   16th KVJ Solutions

2Q) a) Determine the smallest positive integer x, whose last digit is 6 and if we erase this 6 and put it in left most of the number so obtained, the number becomes 4x.

Solution: a) This answer will be posted later.

b) For any real numbers a and b, prove that:

    \(3a^{4}-4a^{3}b+b^{4} \geq 0 \)

Solution: 

\(3a^{4} - 4a^{3}b+b^{4}\geq 0\) for all real numbers a and b .

    =\(a^{4}+2a^{4} +b^{4}- 4a^{3}b\)

    =\(a^{4}+b^{4} +2a^{4} - 4a^{3}b\)

    =\((a^{2})^{2} + (b^{2})^2 + 2a^{4} -4a^{3}b\)

    = \((a^{2})^{2} + (b^{2})^{2} -2a^{2}b^{2} + 2a^{2}b^{2} +2a^{4} - 4a^{3}b \)

    = \((a^{2}-b^{2})^{2} + 2a^{2}b^{2} + 2a^{4}-4a^{3}b\)

    = \((a^{2}-b^{2})^{2}+2a^{2}(b^{2} + a^{2} -2ab)\)

    = \((a^{2}-b^{2})^{2} +2a^{2}(a-b)^{2} \geq 0\) for all a, b

 As square is always positive and sum of the squares also positive.

Hence proved.  

   16th KVJ Solutions


   <<BACK<<                                                 >>NEXT>>      


SHARE YOU ENORMOUS EFFORT AND SMART EXAMPLES HERE

!! NEED MORE HELP !!

SBI! Case Studies