392]let `n` be a positive integer with n≥2. show that n does not divide 2n-1.
ok lets have sum fun in number thoery with some questions. here is first one
1] let x and y be positive integers such that xy divides x^{2}+y^{2}+1 . show that \frac{x^{2}+y^{2}+1}{xy}=3
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1\text{Let } x^2+y^2+1=kxy.
if x=y or y=1,then x=y=1,k=3.
then choose that solution which minimises x+y. wlog let x≥y.
it is easy to get y>1.
then above equation gives the other positive x' satisfying it is,x' =\frac{ky\pm\sqrt{k^{2}y^{2}-4y^{2}-4}}{2}.
We require x'≥x>y.
but xx'=y^2+1. contradiction.
391If n is even then done.
Else (n,2)=1
so, 2n-1≡1(mod n)
so 2n≡2≠1(mod n)
39yup. rite. that was a simple one.
3]show that if n≥6 is composite, then n divides (n-1)!
1the question is self explanatory.
If n is composite, it has two factors p and q,p≠q s.t pq=n and p<n-1,q<n-1. since n≥6.
result follows.
394]suppose that p is a prime number greater than 3. Prove that 7^{p}-6^{p}-1 is divisible by 43.
1i'm a n00b when it comes to number theory(esp the modulus things), are there any good online tutorials on this subject which can be done within a couple of hrs?
39this one is simple
4] find all 3 digit numbers `n` such that `n` is divisible by 11 and \frac{n}{11} is equal to sum of squares of digits of n.
( the same type q was once given in IMO but instead of 11 , it was 5)
1To Vivek:
Try
Weisstein, Eric W. "Congruence." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Congruence.html
Also see AOPSWIKI.
1let the number be abc
case 1.
a+c=b
then 10a+c=a^2+c^2+(a+c)^2. means a+c<9. means a^2+c^2+ac<50. So a<7,c<7.We can manually check the cases remaining.
case2.
a+c=b+11
Again 10(a-1)+c=a^2+c^2+(a+c-11)^2
so 0≤a+c-11<9.
will post abt this later.
395] dtermine the least possible value of natural number `n` such that n ! ends in exactly 1987 zeroes.
15]
n=7960
since there are 1987 zeroes n! is a multiple of 101987 =21987.51987
so we have to find the power of 5 in n!
starting off at a rough guess of 8000 = 1600*5
you get the power of 5 in 8000! as 1600+320+64+12+2 = 1998 ,now thats close!
working backwards at n=1592*5=7960 you get 1987 !!
power of 5 in 7960! = 1592+318+63+12+2=1987
7960!,7961!,7962!,7963!,7964! all end in exactly 1987 zeroes
1@ith_power,thanks for the info,aops was helpful,it had some good explanations instead of generalized crap