[Guest]

We know that, 200 is the smallest positive base ten integer which can not be made prime by changing one of its digits.

M denotes the smallest value of a base N positive integer that cannot changed into a prime by changing a single digit.

For the values of N drawn at random between 4 and 53 inclusively, determine the probability that the first digit of M (reading left to right) is 3.

Note: N is a positive integer and, M cannot contain any leading zero.

Edited May 31, 2010 by K Sengupta

[superprismatic]

The probability that the leading digit is 3 is 0.1 right on the nose!

[Guest]

the 5 out of 50 numbers being:

(200)10 = (3020)4

(200)10 = (310)8

(200)10 = (3 47)51

(200)10 = (3 44)52

(200)10 = (3 41)53

where following is the notation used:

(number)base

Thus, p = 5/50 = 0.1

[Guest]

the 5 out of 50 numbers being:

(200)10 = (3020)4

(200)10 = (310)8

(200)10 = (3 47)51

(200)10 = (3 44)52

(200)10 = (3 41)53

where following is the notation used:

(number)base

Thus, p = 5/50 = 0.1

Check your calculations.

Each of the above results correspond to a conversion of (200)10 to various bases whilst the provisions inclusive of the given problem requires the base N number itself, rather than the base N form of (200)10.

For example, in base 4, (120)4 is the smallest positive integer satisfying the given conditions since each of the base 4 numbers 120, 121, 122 and 123 are composite. But, the first digit in this case is not 3.

Again as an example, it can easily be observed that the duodecimal number 380, that is (380)12 satisfies all the given conditions, including the first digit requirement.

[Guest]

Check your calculations.

Each of the above results correspond to a conversion of (200)10 to various bases whilst the provisions inclusive of the given problem requires the base N number itself, rather than the base N form of (200)10.

For example, in base 4, (120)4 is the smallest positive integer satisfying the given conditions since each of the base 4 numbers 120, 121, 122 and 123 are composite. But, the first digit in this case is not 3.

Again as an example, it can easily be observed that the duodecimal number 380, that is (380)12 satisfies all the given conditions, including the first digit requirement.

If let's say I found a number (let's say decimal) 117 where you can't change any of the digits to make it prime but if you change the left digit to a leading zero it'll become 17 which is a prime, if I found that in one of the bases would it count?

[superprismatic]

The probability that the leading digit is 3 is 0.1 right on the nose!

The 5 are 330₅, 310₈, 380₁₂, 3D0₁₉, and 370₂₀.