For example,

1 = 2^0 * 3^0, r = s = 0

2 = 2^1 * 3^0, r = 1, s = 0

3 = 2^0 * 3^1, r = 0, s = 1

4 = 2^2 * 3^0, r = 2, s = 0

How would you write 5?

23 = 9 + 8 + 6

Actually, we can write, 5 = 2 + 3 = 1 + 4

So, 5 = 2 + 3 = (2^1 * 3^0) + (2^0 * 3^1)

and, 5 = 1 + 4 = (2^0 * 3^0) + (2^2 * 3^0)

Naturally, a formal proof will be needed.

The formal proof: we proceed by induction.

That is to say, we suppose all integers less than n-1 can be represented.

Then we face two cases:

(1) If n is even, then, ...

(2) If n is odd, then, ...

P.S. This problem is originally due to Paul Erdos. Note that the representations need not be unique: for instance,

11 = 2 + 9 = 3 + 8

(2) Find the smallest prime p such that the digits of p (in base 10) add up to a prime number greater than 10.

(2^3) * (3^(-.427816...)) = 5

ReplyDeleteNote that r and s are nonnegative integers

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