Answer to Riddle #8: Remainders

The smallest positive integer that leaves a remainder of 1 when divided by 2, remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and so on up to a remainder of 9 when divided by 10 is 2519.

To find this number, we can use the least common multiple of 2, 3, 4, 5, 6, 7, 8, and 9. This least common multiple is 2520. We then subtract 1 from this number to get 2519. This number is the smallest positive integer that leaves a remainder of 1 when divided by 2, remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and so on up to a remainder of 9 when divided by 10.

Here is a proof by contradiction. Suppose that there is a smaller positive integer that leaves a remainder of 1 when divided by 2, remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and so on up to a remainder of 9 when divided by 10. Let this number be n. Then n must be divisible by the least common multiple of 2, 3, 4, 5, 6, 7, 8, and 9, which is 2520. However, since n is smaller than 2519, it must leave a remainder of 0 when divided by 2520. This is a contradiction, since n leaves a remainder of 1 when divided by 2. Therefore, there is no smaller positive integer that leaves a remainder of 1 when divided by 2, remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and so on up to a remainder of 9 when divided by 10.

The below answer not optimal
We can solve this problem using the Chinese Remainder Theorem, which states that there exists a unique solution (modulo the product of the moduli) to a system of linear congruences with pairwise coprime moduli.

In this case, the system of congruences can be written as follows:

x = 1 (mod 2)
x = 2 (mod 3)
x = 3 (mod 4)
...
x = 9 (mod 10)
Since the moduli are pairwise coprime, the Chinese Remainder Theorem guarantees the existence of a unique solution (modulo 2x3x4x...x10) to this system of congruences.

To find this solution, we can use the following algorithm:

Compute M = 2x3x4x...x10 = 2520.
For each congruence x = a (mod m), compute the value Mi = M/m, and find the inverse of Mi modulo m using the Extended Euclidean Algorithm. Let si be the inverse of Mi modulo m. Compute the solution x as the sum of ai x si x Mi for all i.

Applying this algorithm to our system of congruences, we get:

M = 2x3x4x5x6x7x8x9x10 = 2520
M2 = 3x4x5x6x7x8x9x10 = 181440
M3 = 2x4x5x6x7x8x9x10 = 90720
M4 = 2x3x5x6x7x8x9x10 = 63504
M5 = 2x3x4x6x7x8x9x10 = 50400
M6 = 2x3x4x5x7x8x9x10 = 43200
M7 = 2x3x4x5x6x8x9x10 = 38720
M8 = 2x3x4x5x6x7x9x10 = 36288
M9 = 2x3x4x5x6x7x8x10 = 35280

Using the Extended Euclidean Algorithm, we find the following values for the inverses:

s2 = 4
s3 = 5
s4 = 7
s5 = 5
s6 = 5
s7 = 23
s8 = 7
s9 = 23

Finally, we can compute the solution as follows:

x = (1 x s2 x M2 + 2 x s3 x M3 + 3 x s4 x M4 + 4 x s5 x M5 + 5 x s6 x M6 + 6 x s7 x M7 + 7 x s8 x M8 + 8 x s9 x M9) modulo M

Plugging in the values, we get:

x = (1 x 4 x 181440 + 2 x 5 x 90720 + 3 x 7 x 63504 + 4 x 5 x 50400 + 5 x 5 x 43200 + 6 x 23 x 38720 + 7 x 7 x 36288 + 8 x 23 x 35280) modulo 2520
x = 2321 modulo 2520

Therefore, the smallest positive integer that leaves a remainder of 1 when divided by 2, remainder of 2 when divided by 3, a remainder

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