Search found 72 matches

by zrosnbrick
Mon Apr 23, 2007 4:34 am
Forum: Number
Topic: 7^a + 1 = 2^b
Replies: 2
Views: 1664

I'll post my solution, though I don't like it much. Consider that (0, 1) and (1, 3) are the only solutions with a [le] 1 and b [le] 3. Now, for any solution, 7 a + 1 would have to be a multiple of 16: 7 a + 1 [cong] 0 (mod 16) [equiv] (8 - 1) a + 1 [cong] 0 (mod 16) [equiv] [sum]nCr( a , k )(8 k )((...
by zrosnbrick
Sun Apr 22, 2007 7:47 pm
Forum: Number
Topic: 7^a + 1 = 2^b
Replies: 2
Views: 1664

7^a + 1 = 2^b

Find all solutions in nonnegative integers to 7a + 1 = 2b.
by zrosnbrick
Wed Apr 11, 2007 9:15 pm
Forum: Number Theory
Topic: a divisibility criterion
Replies: 2
Views: 1739

Consider the number a . For every odd a not divisible by 5, there is a number n that multiplied by a ends in a one. More precisely consider the table where the first column is the last digit of a and the second column is n : 1 [map] 1 3 [map] 27 7 [map] 3 9 [map] 9 So now multiplying a by n will res...
by zrosnbrick
Wed Apr 04, 2007 12:48 am
Forum: Geometry
Topic: Triangle Ratio
Replies: 1
Views: 2545

First draw in the line segment connecting C and M . Let [ P 1 P 2 ... P n ] be the area of polygon P 1 P 2 ... P n . WLOG let [ ABC ] = 1, and let a = [ BDM ] , b = [ BAM ] , c = [ AME ] , d = [ EMC ] and e = [ MCD ] . Since DM = MA , a = b [imp] 2 a = a + b Since 2 BD = DC , 2 a = e and 2( a + b )...
by zrosnbrick
Wed Apr 04, 2007 12:04 am
Forum: Geometry
Topic: Triangle Ratio
Replies: 1
Views: 2545

Triangle Ratio

http://zefrosnbrick.com/Picture%201.png In triangle ABC , lines AD and BE are drawn, where point D lies on side BC and point E lies on side AC . These lines meet at M , which is the midpoint of AD . If BD = ( 1 / 3 ) BC , find (with proof) the fraction of the area of the triangle ABC that is occupi...
by zrosnbrick
Tue Apr 03, 2007 9:41 pm
Forum: Number
Topic: Sums of Powers of Real Numbers
Replies: 10
Views: 3508

I think you guys are looking at it the wrong way... There is not extra constraint, it's just that that a's are all unknown constants - we don't know how many are positive.
by zrosnbrick
Tue Apr 03, 2007 5:59 pm
Forum: Number
Topic: Sums of Powers of Real Numbers
Replies: 10
Views: 3508

Sums of Powers of Real Numbers

Let a1, a2, ..., a10 be ten real numbers whose product is positive. Prove that there is some odd positive integer k such that (a1)k + (a2)k + ... + (a10)k is nonzero.


This problem seems tantalizingly trivial, and yet I can't seem to get it.
by zrosnbrick
Tue Mar 13, 2007 2:06 am
Forum: Number
Topic: The (impossible) 1994 number challenge!
Replies: 8
Views: 3524

Here is my mostly complete solution using an adaptation of the program I wrote for the euler project problem: 1 = (1 + 9 - 9) 4 2 = (1 + 9) / (9 - 4) 3 = 1 + 9 - 9 + [radic]4 4 = (1 + 9 - 9) * 4 5 = 1 + 9 - 9 + 4 6 = 1 9 + 9 - 4 7 = 1 - 9 - 9 + 4! 8 = (1 + 9 / 9) * 4 9 = (1 * 9 + 9) / [radic]4 10 = ...
by zrosnbrick
Sun Mar 11, 2007 3:34 am
Forum: Number
Topic: sequence
Replies: 11
Views: 4129

How about the sum of the first n squares?

[sum]k2 = (n)(n + 1)(2n + 1) / 6

P.S. Try searching Sloane's next time.
by zrosnbrick
Sat Mar 10, 2007 9:19 pm
Forum: Number
Topic: Three squares make a sixth power
Replies: 8
Views: 3491

Quintic: Can you show how you arrived at your factorization?
by zrosnbrick
Sat Mar 10, 2007 4:30 am
Forum: Number
Topic: Three squares make a sixth power
Replies: 8
Views: 3491

n 6 = a 2 + b 2 - c 2 => n 6 - a 2 = b 2 - c 2 => (n 3 + a)(n 3 - a) = (b + c)(b - c) Let n 3 - a = 3 => a = n 3 - 3, n 3 > 3 => n > 1 (n 3 + n 3 - 3)(3) = (b + c)(b - c) => 6n 3 - 9 = (b + c)(b - c) Let b - c = 1 => c = b - 1 => 6n 3 - 9 = 2b - 1 => 6n 3 - 8 = 2b => b = 3n 3 - 4 => c = 3n 3 - 5 We ...
by zrosnbrick
Sun Feb 04, 2007 5:37 am
Forum: Number Theory
Topic: Analytical Number Problem (2)
Replies: 3
Views: 2326

A somewhat elegant solution WLOG, A [le] B [le] C [imp] 1 / A [ge] 1 / B [ge] 1 / C [imp] 1 / A [ge] ( 1 / 3 ) * ( 7 / 15 ) However, 1 / A must also be less than 7 / 15 so 7 / 45 [le] 1 / A < 7 / 15 [imp] 3 [le] A [le] 6 Subtracting the four different values of A, we get [tt] [/tt]A = 3 [imp] 1 / B ...
by zrosnbrick
Sun Jan 28, 2007 1:45 am
Forum: Number Theory
Topic: A Planet Problem
Replies: 7
Views: 3142

No, because if the astronomer on planet A is looking at planet B, then obviously the astronomer on planet B will be looking at planet A because if A is closest to B then B is in turn closest to A, and since there are an odd number of planets, one will always be left out.
by zrosnbrick
Tue Dec 26, 2006 7:10 pm
Forum: Number Theory
Topic: smallest multiple of 17 (urgent! )
Replies: 6
Views: 3732

This isn't fast, but consider approaching the problem like this: We want to multiply 17 by a one digit number so that it ends in one one: a * 17 (mod 10) [cong] 1 [imp] a = 3, now for a two digit number that ends in two ones, which must now end in 3 (10a + 3) * 17 (mod 100) [cong] 11 [imp] a = 8, no...
by zrosnbrick
Tue Dec 26, 2006 6:31 am
Forum: Number Theory
Topic: smallest multiple of 17 (urgent! )
Replies: 6
Views: 3732

I wrote a simple computer program and found a solution when n=16.
by zrosnbrick
Tue Dec 26, 2006 1:53 am
Forum: Number Theory
Topic: 9^105
Replies: 4
Views: 3792

Consider that 9 105 = (10 - 1) 105 . From the binomial coefficients theorem, (10 - 1) 105 = [sum]nCr(105, k) * 10 k * (-1) 105 - k , k = 0, 105. If we were to take this number modulo 1000, we could exclude every value of k greater than 2 because it would yield a number that is a multiple of 1000. Th...
by zrosnbrick
Mon Dec 25, 2006 4:38 pm
Forum: Number Theory
Topic: extremely urgent help please!
Replies: 2
Views: 2160

Consider that if n is represented in its prime factorization p a q b r c ... the number of divisors it has is (a + 1)(b + 1)(c + 1)... (for a proof of this see here ). Prime factorizing 75, we see that 75 = 3 1 5 2 and so 75 has exactly (1 + 1)(2 + 1) = 6 factors. These can be verified as: 1, 3, 5, ...
by zrosnbrick
Fri Dec 15, 2006 11:49 pm
Forum: Combinatorics
Topic: Chance in mod
Replies: 3
Views: 2872

Consider the multiplication table (mod 9) 0 1 2 3 4 5 6 7 8 0 0 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 7 8 2 0 2 4 6 8 1 3 5 7 3 0 3 6 0 3 6 0 3 6 4 0 4 8 3 7 2 6 1 5 5 0 5 1 6 2 7 3 8 4 6 0 6 3 0 6 3 0 6 3 7 0 7 5 3 1 8 6 4 2 8 0 8 7 6 5 4 3 2 1 So the chances for each congruence is: 0: 21 / 81 = 26% 1: 6...
by zrosnbrick
Thu Nov 23, 2006 2:25 am
Forum: Number Theory
Topic: prove FLT for n=3
Replies: 1
Views: 1667

Look here.
by zrosnbrick
Sun Nov 05, 2006 1:55 am
Forum: Number
Topic: Fibonacci Infinite Sum
Replies: 2
Views: 2217

Fibonacci Infinite Sum

Find the sum from n=1 [map] n=[inf] of Fn / 3n where Fn is the nth Fibonacci number.