maths questions revised?

Question

1a) Find all integers >1 for which n^4+4^n is a prime.
b) Show that 1001| (1^1993+2^1993+...+1000^1993)
c) 2903^n-803^n-464^n+261^n is divisible by 1897 for all natural numbers n.

2) Find the remained when the sum 1+2+3...+100 is divided by 4.

3) Find the remained der when 1+2!+3!...+100! is divided by 12.

4) Are there any integers x and y such that x^2-5y^2=2?
b) How many integers are there such that n such that 2^n+27 is divisible by 7?
5) Is it true that every year including any leap year, has at least one Friday 13th? Justify your answer.

Answer 1

#1a: There aren't any. Consider the following:

n^4 + 4^n = n^4 + 2n^2*2^n + 4^n - 2n^2*2^n = (n^2 + 2^n)^2 - 2n^2*2^n = (n^2 + 2^n)^2 - n^2*2^(n+1) = (n^2 + 2^n - n * 2^((n+1)/2)) * (n^2 + 2^n + n * 2^((n+1)/2)). Therefore, if n is odd, n^2 + 2^n - n * 2^((n+1)/2) is an integer factor of n^4 + 4^n. It can be shown that if n > 1, this factor is not equal to 1 or the number, therefore if n is odd, n^4 + 4^n is composite. Further, if n is even, n^4 + 4^n is even and therefore composite. Thus, there are no integers greater than 1 for which n^4 + 4^n is prime. Q.E.D.

#1b: Note that 1^1993 + 2^1993 ... 1000^1993 = 1^1993 + 1000^1993 + 2^1993 + 999^1993 ... 500^1993 + 501^1993 ≡ 1^1993 + (-1)^1993 + 2^1993 + (-2)^1993... mod 1001. Since we are raising these numbers to an odd exponent, we factor out the -1 to get 1^1993 - 1^1993 + 2^1993 - 2^1993... 500^1993 - 500^1993 mod 1001, which is obviously zero. Thus, 1001 divides the sum 1^1993 + 2^1993... 1000^1993. Q.E.D.

#1c: Consider a polynomial x^n - c^n for any constant c. Clearly, this polynomial has a root at x=c, and thus by the fundamental theorem of algebra, (x-c) must be a factor of it. Thus, 2903^n-803^n-464^n+261^n = (2903-803)(something) - (464-261)(something else). However, 2903-803 ≡ 0 mod 7 and 464-261 ≡ 0 mod 7, thus 2903^n-803^n-464^n+261^n ≡ 0 - 0 ≡ 0 mod 7. Similarly, 2903^n-803^n-464^n+261^n = 2903^n-464^n-803^n+261^n = (2903 - 464)(something) - (803-261)(something else) ≡ 0-0 ≡ 0 mod 271. Thus, 2903^n-803^n-464^n+261^n is divisible by both 7 and 271, and must therefore be divisible by the LCM of 7 and 271, but LCM (7, 271) = 1897. Therefore 2903^n-803^n-464^n+261^n is divisible by 1897 for all n. Q.E.D.

And you sir, are a bastard for lumping those three problems together as though they had a common solution method. I'm afraid that my answers to a and c were produced only through the liberal use of google (though I did get b on my own). The remaining problems are, fortunately, much easier.

#2: 5050 ≡ 2 mod 4. I've heard the story about Gauss so many times that I've just memorized the value of (n=1, 100)∑n.

#3: 9. Note that 12|4!, and 4! divides n! for n>4, so the only terms which are not ≡ 0 mod 12 are 1, 2, and 6.

#4a: No. 5y^2≡0 mod 5, and x^2≡0, 1, or 4 mod 5, thus x^2-5y²≡0, 1, or 4 mod 5, and 2 is not on that list.

#4b: Infinitely many: specifically, every multiple of 3. Note 2^(3n) = 8^n ≡ 1 mod 7, and 1+27 ≡ 0 mod 7.

#4c: Yes. A month contains a friday the 13th iff it starts on a Sunday. Let the integers mod 7 be the days of the week and 0 be the day January starts on. Then the months start on the days [0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5] during a normal year, or [0, 3, 4, 0, 2, 5, 0, 3, 6, 1, 4, 6] on leap years. In either case, for each day of the week, there is at least one month that starts with that day, therefore there is at least one month that starts with Sunday, and therefore at least one Friday the 13th.

Answer 2

even as your son says the "more effective than" signal might want to be grew to grow to be into an equals it really is the most important to fixing those inequalities. you may want to do not ignore that y=mx+c the position m and c are any numbers supplies the equation of a immediately line. on your get at the same time replacing the more effective than with an equals signal supplies y - 2x = a million or y = 2x + a million. This immediately line divides the x-y airplane into 2 elements. in a unmarried area each and each of the (x,y) values fulfill y > 2x + a million and in the different they fulfill y < 2x +a million. on the line itself we've y = 2x + a million. checking out the 4 factors supplies a) 2*3+a million = 7 and a million < 7 b) 2*a million+a million = 3 and three = 3 c) 2*0 +a million= a million and a million = a million d) 2*0 +a million =a million and a million> -a million. i visit't see in case your inequality signal is more effective than or equivalent to, yet when it is in uncomplicated words more effective than then not one of the criteria are in the answer set. If the inequality signal contains an equals then b) and d) are in the answer set.

Answer 3

What you're doing here -- asking for others to do your whole assignment for you -- is considered cheating. You are missing the opportunity to practice these concepts, and you are also being dishonest.

My advice is to reread your notes and your textbook, then try the problems, as best you can. Then, if there is a specific point you don't understand, ask for help and you'll get more responses.

Answer 4

If you don't work problems and *do* the homework, you won't learn the subject.


Doug