Powers & Permutation of Digits

Prove that the list of the numbers with the property described below is finite.

I claim that you cannot find a k-digit number (k ≥ 12) with this property.

N.B. There are less than 50,000 which have your property (in base 10)

CASE #1 :

1364 & 6143 : The digits of 1,3,6,4 raised to the power of digits 6,1,4,3.
6143 is a permutation of 1364.

(1^6) + (3^1) + (6^4) + (4^3) = 1364

4316 & 6143 :

(4^6) + (3^1) + (1^4) + (6^3) = 4316

1364 and 4316 are permutations of 6143

4355 & 5435 :

(4^5) + (3^4) + (5^3) + (5^5) = 4355

067236 & 326706 :

(3^0) + (2^6) + (6^7) + (7^2) + (0^3) + (6^6) = 326706

Other examples: 15630, 17463, 48625, 38650

CASE #2 : The digits & powers representing the digits one number give a permutation of that number.

3435 & 5343 :

(5^5) + (3^3) + (4^4) + (3^3) = 3435

CASE #3 :

4096 = 4^6 + 0^9

The digits 4 and 0 are raised to powers 6 and 9, and all these digits represent the digits of the number

Other examples,
397612 = 3^2 + 9^1 + 7^6 + 6^7+ 1^9+ 2^3
48625 = 4^5 + 8^2 + 6^6 + 2^8 + 5^4

Advertisements

About benvitalis

math grad - Interest: Number theory
This entry was posted in Math Beauty and tagged , . Bookmark the permalink.

1 Response to Powers & Permutation of Digits

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s