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MATHMAA

# 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