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Calling all positive integers ...

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Posted · Report post

OK maybe not all.

How many pairs of mutually prime positive integers sum to 2x10250?

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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Posted · Report post

4x10

249

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Posted · Report post

4x10249-1, here's why:



The pairs must contain only odd numbers, so this brings us down to 10250 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)x10250 which equals 8x10249. 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 8x249-2 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 4x10249-1 pairs.
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Posted · Report post

Agree with Superprismatic. I think u have counted (1, 2*10^250-1) also.

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Posted · Report post

4x10

249-1, here's why:

The pairs must contain only odd numbers, so this brings us down to 10250 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)x10250 which equals 8x10249. 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 8x249-2 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 4x10249-1 pairs.

Consider the case of 2 x 101. Answer: 4 x 100.

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 2x10250 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)?

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Posted · Report post

4x10

249-1, here's why:

The pairs must contain only odd numbers, so this brings us down to 10250 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)x10250 which equals 8x10249. 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 8x249-2 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 4x10249-1 pairs.

Consider the case of 2 x 101. Answer: 4 x 100.

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 2x10250 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!

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