July 2023
Problem of the Month

Ordered Triples Problem
Submitted by Brijesh Dave

How many ordered triples of integers
(a, b, c), with a ≥ 2, b ≥ 1, and c ≥ 0,
satisfy both log_ab = c²⁰⁰⁵ and
a + b + c = 2005?

Solution:

There are two ordered triples (2004, 1, 0) and (1002, 1002, 1).

[EXPLANATION]:
Use the definition of a logarithm:
If log_xy = w, then y = x^w


Look at different possibilities for c:

If c = 0, then a⁰ = b.   therefore b = 1.
    So, a = 2005 - 1 - 0 = 2004.
    So there is one triple: (2004, 1, 0)
    that satisfies the conditions.

If c = 1, then a¹ = b.   therefore b = a.
    So, a + a + 1 = 2005, so a = 1002
    and b = 1002.
    So, (1002, 1002, 1) is another solution.

If c ≥ 2, the exponent of a becomes very large, and since a ≥ 2, it is impossible for three numbers a, b, and c to add up to 2005.

Correctly solved by:

1. K.Sengupta	Calcutta, INDIA
2. Davit Banana	Istanbul, Turkey
3. Carlos de Armas	Barcelona City, Spain
4. Seth Cohen	Concord, New Hampshire
5. Rob Miles	Northbrook, Illinois

Send any comments or questions to: David Pleacher