The numbers 637, 638, and 639 constitute a set of three consecutive positive integers (in order) that are respectively divisible by 13, 11, and 9.
Find the first set of four consecutive positive integers (in order) that are respectively divisible by 13, 11, 9, and 7.
How about the first set of five consecutive positive integers (in order) that are respectively divisible by 13, 11, 9, 7, and 5?
Lets call the consecutive numbers x, x+1, x+2, ... etc.
Then I'll start with just one congruence:
x mod 13 = 0.
This has a trivial solution of x being any multiple of 13.
Given x = 0 mod 13, add the second congruence:
x+1 mod 11 = 0, or equivalently x = 10 mod 11.
Then apply the Chinese Remainder Theorem to get x = 65 mod 143.
Given x = 65 mod 143, add the third congruence:
x+2 mod 9 = 0, or equivalently x = 7 mod 9.
Then apply the Chinese Remainder Theorem to get x = 637 mod 1387.
Now we can generate the example triplet 637, 638, and 639 being respectively divisible by 13, 11, and 9.
Given x = 637 mod 1287, add the fourth congruence:
x+3 mod 7 = 0, or equivalently x = 4 mod 7.
Then apply the Chinese Remainder Theorem to get x = 4498 mod 9009.
Now we can generate the answer to the first question 4498, 4499, 4500, and 4501 being respectively divisible by 13, 11, 9, and 7.
Given x = 4498 mod 9009, add the fifth congruence:
x+4 mod 5 = 0, or equivalently x = 1 mod 5.
Then apply the Chinese Remainder Theorem to get x = 22516 mod 45045.
Now we can generate the answer to the second question 22516, 22517, 22518, 22519, and 22520 being respectively divisible by 13, 11, 9, 7, and 5.
Solution
