bonanova 77 Report post Posted May 6, 2013 OK maybe not all. How many pairs of mutually prime positive integers sum to 2x10^{250}? You should be able to do this on the back of an envelope, and probably can't count the pairs with a simple program (before the earth freezes.) Pairs differing only in the order of addition count as 1 pair, not as 2 different pairs. Share this post Link to post Share on other sites

0 witzar 18 Report post Posted May 6, 2013 4x10^{249} Share this post Link to post Share on other sites

0 superprismatic 11 Report post Posted May 6, 2013 4x10^{249}-1, here's why: The pairs must contain only odd numbers, so this brings us down to 10^{250} possible pairs. Furthermore, each pair must contain only those odd numbers which are not equal to 0 modulo 5. This brings us to (4/5)x10^{250} which equals 8x10^{249}. But, this is 2 too many because the sequence of odd numbers begins (modulo 5) with 1,3,2,4,1,3,2,4... but ends in 1,3 which is the middle of this 4-long cycle. Which brings us to 8x^{249}-2 pairs. But this is double what we want because it counts each pair twice, counting the order of addition both ways. Therefore, we need to divide this in two to get 4x10^{249}-1 pairs. Share this post Link to post Share on other sites

0 dark_magician_92 4 Report post Posted May 6, 2013 Agree with Superprismatic. I think u have counted (1, 2*10^250-1) also. Share this post Link to post Share on other sites

0 bonanova 77 Report post Posted May 7, 2013 4x10^{249}-1, here's why: The pairs must contain only odd numbers, so this brings us down to 10^{250} possible pairs. Furthermore, each pair must contain only those odd numbers which are not equal to 0 modulo 5. This brings us to (4/5)x10^{250} which equals 8x10^{249}. But, this is 2 too many because the sequence of odd numbers begins (modulo 5) with 1,3,2,4,1,3,2,4... but ends in 1,3 which is the middle of this 4-long cycle. Which brings us to 8x^{249}-2 pairs. But this is double what we want because it counts each pair twice, counting the order of addition both ways. Therefore, we need to divide this in two to get 4x10^{249}-1 pairs. Consider the case of 2 x 10^{1}. Answer: 4 x 10^{0}. 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 11 10 10 Suppose I were to argue that 2x10^{250} constructs an integral number of similar groups of 10 sums, each of which gives back 6 rows to the mutually prime condition. Which group contributes the (-1)? Share this post Link to post Share on other sites

0 superprismatic 11 Report post Posted May 7, 2013 4x10^{249}-1, here's why: The pairs must contain only odd numbers, so this brings us down to 10^{250} possible pairs. Furthermore, each pair must contain only those odd numbers which are not equal to 0 modulo 5. This brings us to (4/5)x10^{250} which equals 8x10^{249}. But, this is 2 too many because the sequence of odd numbers begins (modulo 5) with 1,3,2,4,1,3,2,4... but ends in 1,3 which is the middle of this 4-long cycle. Which brings us to 8x^{249}-2 pairs. But this is double what we want because it counts each pair twice, counting the order of addition both ways. Therefore, we need to divide this in two to get 4x10^{249}-1 pairs. Consider the case of 2 x 10^{1}. Answer: 4 x 10^{0}. 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 11 10 10 Suppose I were to argue that 2x10^{250} constructs an integral number of similar groups of 10 sums, each of which gives back 6 rows to the mutually prime condition. Which group contributes the (-1)? It seems rather difficult to pull the wool over the Denizens! Share this post Link to post Share on other sites

OK maybe not all.

How many pairs of mutually prime positive integers sum to 2x10

^{250}?You should be able to do this on the back of an envelope, and probably can't count the pairs with a simple program (before the earth freezes.)

Pairs differing only in the order of addition count as 1 pair, not as 2 different pairs.

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