perplexus.info :: Algorithms : Divisible by Prime Power

Divisible by Prime Power (Posted on 2023-05-15)

Define d(x) as sum of the digits of x, where x is a hexadecimal positive integer.
d(d(x)) denotes the sum of digits of d(x).

For example, when x=(ABC)₁₆
Then, d(x) = (A)₁₆+(B)₁₆+(C)₁₆ = (21)₁₆
and, d(d(x)) = 2 + 1 = 3

Consider the first 1011 (base ten) values of a hexadecimal prime number N.

Devise an algorithm such that:
• Each of d(N) and d(d(N)) is divisible by a prime power.
Note: A prime power is a number of the form pⁿ, where p is a prime number, and n is an integer greater than 1.

Submitted by K Sengupta

Rating: 5.0000 (1 votes)

Solution: (Hide)

The first 1011 hexadecimal primes do not have the desired property.
Checking for 1029 (dec) which is 2011(hex) to 4998(dec) which is BDCD (hex), we find that 2312nd prime is equal to 20479(dec) which is equal to 4FFF (hex), so the digits add to 4+3*15 = 49, which is 7^2, and is represented in hex as 31, which adds up to 4, the square of 2.
For a full explanation, refer to the solution submitted by Charlie in this location.

Comments: ( You must be logged in to post comments.)

	Subject	Author	Date
	computer solution	Charlie	2023-05-15 13:57:09

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