January 7, 2013
Geometry: A given convex polygon has d diagonals (lines other than sides
which connect vertices). By adding one more side you add 10 diagonals.
How many sides does the polygon now have?
From NCTM
Answer available January 14, next puzzle
Answer to math puzzle of December 31
Sequences: How many sequences of consecutive positive integers add to 2013?
From NCTM
Solution: For any particular integer there are two types of solutions: Those with k terms where k is an odd factor of the number, and those with 2 terms when the number is odd. The latter for 2013 is 1006+1007. The others are for the odd factors of 2013: 3,11,33, and 61; 3 terms starting with 670, 11 terms starting with 178, 33 starting with 45, and 61 starting with 3. There are more, longer sequences of integers (183 and 671), but they include negative integers. Total of 5 sequences. ( It follows that no power of 2 is equal to the sum of a sequence of integers!)
Given any number n, to find the sequence of length k (a factor of n), divide by k, subtract (k-1)/2 to get the first number in the sequence. So above, n=2013, k=11, get 183 minus 5 so the sequence of 11 integers starting with 178 adds to 2013.
©copyright 2013, Louis Bookbinder - booky1@earthlink.net
updated 7 January 2013
From NCTM
Answer available January 14, next puzzle
Answer to math puzzle of December 31
Sequences: How many sequences of consecutive positive integers add to 2013?
From NCTM
Solution: For any particular integer there are two types of solutions: Those with k terms where k is an odd factor of the number, and those with 2 terms when the number is odd. The latter for 2013 is 1006+1007. The others are for the odd factors of 2013: 3,11,33, and 61; 3 terms starting with 670, 11 terms starting with 178, 33 starting with 45, and 61 starting with 3. There are more, longer sequences of integers (183 and 671), but they include negative integers. Total of 5 sequences. ( It follows that no power of 2 is equal to the sum of a sequence of integers!)
Given any number n, to find the sequence of length k (a factor of n), divide by k, subtract (k-1)/2 to get the first number in the sequence. So above, n=2013, k=11, get 183 minus 5 so the sequence of 11 integers starting with 178 adds to 2013.
©copyright 2013, Louis Bookbinder - booky1@earthlink.net
updated 7 January 2013