Four three-digit numbers are in arithmetic progression, and the number of factors of each is also in arithmetic progression. In fact that second arithmetic progression has a constant difference of 1, so that each successive number in the original arithmetic progression has one more factor than the number before it.
Note: These are not the prime factors of the numbers, but rather any factor, including the number itself and 1, so, for example, 46 has four factors (1, 2, 23 and 46), as does 8 (1, 2, 4 and 8).
What is the original arithmetic progression of three-digit numbers?
(In reply to
Solution / method by ed bottemiller)
"...Clearly a task for a computer program. " ????
Not at all !!
I have solved it in less than 20 minutes, using pen and paper and a short list of three digit squares (121,144,....961)
1. It easy to show that only a square number has an odd number of divisors.
2. 2,3,4,5 is the most logical choice.
3. 3 and 5 correspond to squares of the same parity , since those squares differ by 2*d and d has to be an integer
4. 3 divisors define a square of a prime , thus limiting the eligibility for the second number to only 7 ODD candidates: 121,169,289,361,529,841 and 961
5. The 4th number had to be 625=5^4 (only a 4th power of a prime has exactly 5 divisors) : 3^4 too small; 7^4 too big.
6.Testing 361 produced 2*d=264 and therefore the AP with d=132. Ergo: 229, 361, 493, and 625
ALL THE ABOVE RESULTS FROM THE FORMULA:
N -the number of divisors of (2^A)*(3^B)*(5^C)*... (P^K)is equal to (A+1)*(B+1)*(C+1)*...(K+1)*
re: Solution / method ==>>MY WAY
