# 1922. Count Good Numbers

## Description

A digit string is good if the digits (0-indexed) at even indices are even and the digits at odd indices are prime (2, 3, 5, or 7).

• For example, "2582" is good because the digits (2 and 8) at even positions are even and the digits (5 and 2) at odd positions are prime. However, "3245" is not good because 3 is at an even index but is not even.

Given an integer n, return the total number of good digit strings of length n. Since the answer may be large, return it modulo 109 + 7.

A digit string is a string consisting of digits 0 through 9 that may contain leading zeros.

Example 1:

Input: n = 1
Output: 5
Explanation: The good numbers of length 1 are "0", "2", "4", "6", "8".


Example 2:

Input: n = 4
Output: 400


Example 3:

Input: n = 50
Output: 564908303


Constraints:

• 1 <= n <= 1015

## Solutions

• class Solution {
private int mod = 1000000007;

public int countGoodNumbers(long n) {
return (int) (myPow(5, (n + 1) >> 1) * myPow(4, n >> 1) % mod);
}

private long myPow(long x, long n) {
long res = 1;
while (n != 0) {
if ((n & 1) == 1) {
res = res * x % mod;
}
x = x * x % mod;
n >>= 1;
}
return res;
}
}

• int MOD = 1000000007;

class Solution {
public:
int countGoodNumbers(long long n) {
return (int) (myPow(5, (n + 1) >> 1) * myPow(4, n >> 1) % MOD);
}

private:
long long myPow(long long x, long long n) {
long long res = 1;
while (n) {
if ((n & 1) == 1) {
res = res * x % MOD;
}
x = x * x % MOD;
n >>= 1;
}
return res;
}
};

• class Solution:
def countGoodNumbers(self, n: int) -> int:
mod = 10**9 + 7

def myPow(x, n):
res = 1
while n:
if (n & 1) == 1:
res = res * x % mod
x = x * x % mod
n >>= 1
return res

return myPow(5, (n + 1) >> 1) * myPow(4, n >> 1) % mod


• const mod int64 = 1e9 + 7

func countGoodNumbers(n int64) int {
return int(myPow(5, (n+1)>>1) * myPow(4, n>>1) % mod)
}

func myPow(x, n int64) int64 {
var res int64 = 1
for n != 0 {
if (n & 1) == 1 {
res = res * x % mod
}
x = x * x % mod
n >>= 1
}
return res
}