Problem Solving

ajith wrote:

sanju09 wrote: Isn't that p^2 - 5 = ( p^2 - 1) - 4?

Rest is fabulous!!

I was about to edit the post.. you caught me before that..

that look is not cool

keep looking like only

ajith wrote:

sanju09 wrote:
If 5 + 3 n were the perfect square of some integer, say p, then

5 + 3 n = p^2

3 n = p^2 - 5

But, p^2 - 5 is NEVER divisible by 3. WHY?

Because

p^2-5=( p^2-1) +4
= ( p+1) (p-1) +4

Now p-1, p, p+1 are three consecutive integers , one of them should be a multiple of 3

If p-1 or p+1 is a multiple of 3; p^2-5 is not because 4 is not a multiple of 3
if p is a multiple of 3 p^2 is a multiple of 3 but p^2 -5 is not a multiple of 3

When p is a multiple of 3, then we can be certain that p^2 - 5 is NOT a multiple of 3; very well! But when it's so, neither p + 1 nor p - 1 is a multiple of 3. Then how can we make it certain that (p + 1) (p - 1) ± 4 is not a multiple of 3? Ajith? There are few more cards left to be dealt.

sanju09 wrote: When p is a multiple of 3, then we can be certain that p^2 - 5 is NOT a multiple of 3; very well! But when it's so, neither p + 1 nor p - 1 is a multiple of 3. Then how can we make it certain that (p + 1) (p - 1) ± 4 is not a multiple of 3? Ajith? There are few more cards left to be dealt.

(p + 1) (p - 1) - 4 = p^2 - 5 ; we can either look at p^2 - 5 or at (p + 1) (p - 1) - 4 ; both are the same. so if p^2 - 5 is not a multiple of 3, neither is (p + 1) (p - 1) - 4

I think that will suffice, No cards unturned here!

ajith wrote:

sanju09 wrote: When p is a multiple of 3, then we can be certain that p^2 - 5 is NOT a multiple of 3; very well! But when it's so, neither p + 1 nor p - 1 is a multiple of 3. Then how can we make it certain that (p + 1) (p - 1) ± 4 is not a multiple of 3? Ajith? There are few more cards left to be dealt.

(p + 1) (p - 1) - 4 = p^2 - 5 ; we can either look at p^2 - 5 or at (p + 1) (p - 1) - 4 ; both are the same. so if p^2 - 5 is not a multiple of 3, neither is (p + 1) (p - 1) - 4

I think that will suffice, No cards unturned here!

I learnt all that when p is a multiple of 3, and agreed too. But, what if p is not so. Will that still work? Turn Jokers NOW!

sanju09 wrote: I learnt all that when p is a multiple of 3, and agreed too. But, what if p is not so. Will that still work? Turn Jokers NOW!

If p is not a multiple of 3, either p-1 or p+1 will be a multiple of 3 and hence (p + 1) (p - 1) will be a multiple of 3 ; hence (p + 1) (p - 1) - 4 will not be a multiple of 3; p^2-5 will not be a multiple of 3.

Told ya, I am counting my cards!
Joker is over! Dark Knight killed him!

ajith wrote:

sanju09 wrote: I learnt all that when p is a multiple of 3, and agreed too. But, what if p is not so. Will that still work? Turn Jokers NOW!

If p is not a multiple of 3, either p-1 or p+1 will be a multiple of 3 and hence (p + 1) (p - 1) will be a multiple of 3 ; hence (p + 1) (p - 1) - 4 will not be a multiple of 3; p^2-5 will not be a multiple of 3.

Told ya, I am counting my cards!
Joker is over! Dark Knight killed him!

hmmmm Bright Knight, do not turn Joker please

ajith wrote:

sanju09 wrote:
If 5 + 3 n were the perfect square of some integer, say p, then

5 + 3 n = p^2

3 n = p^2 - 5

But, p^2 - 5 is NEVER divisible by 3. WHY?

Because

p^2-5=( p^2-1) +4
= ( p+1) (p-1) +4

Now p-1, p, p+1 are three consecutive integers , one of them should be a multiple of 3

If p-1 or p+1 is a multiple of 3; p^2-5 is not because 4 is not a multiple of 3
if p is a multiple of 3 p^2 is a multiple of 3 but p^2 -5 is not a multiple of 3

ok thats a night solution ajith except for p^2-5=( p^2-1) +4,which u corrected. I read this type of solution by you in one more post...and it is really nice you are thinking in such a splendid way.. you will be solving the problem very fast..

sanju09 wrote:

shashank.ism wrote:

sanju09 wrote:For some integer n, 5 + 3 n is NEVER
(A) a prime
(B) divisible by 7
(C) 1 in magnitude
(D) a perfect cube
(E) a perfect square

A 5+6 =11 is prime
B 5+ 9 = 14 divisble by 7
C 5-6 = -1 mag = 1
D 5+3 = 8 a perfect cube
E left option is E so I go with E is there a way to check E..

If 5 + 3 n were the perfect square of some integer, say p, then

5 + 3 n = p^2

3 n = p^2 - 5

But, p^2 - 5 is NEVER divisible by 3. WHY?

There are only two resolutions for p, either it's PRIME to 3 or NOT. In the first case we can safely take p^2 - 1 as some multiple of 3, say 3 m (m is a positive integer). Now, p^2 - 5 = p^2 - 1 - 3 - 1 = 3 m - 3 - 1 = 3 (m - 1) - 1, and 3 CANNOT divide -1. In the second case, when p is not prime to 3, then p is a multiple of 3, let's again say p^2 = 3 m so that p^2 - 5 = p^2 - 6 + 1= 3 m - 6 + 1 = 3 (m - 2) + 1, and again, 3 CANNOT divide +1 either.