Compute the smallest positive integer k such that 49 divides (k2k).
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
The largest a such that 7a(k2k) is equal to the number of carries when you add k+k in base 7 , by Kummer's Theorem. Thus, we need two carries, so 2k must have at least 3 digits in base 7 . Hence, k≥25. We know k=25 works because 25+25=347+347=1017 has two carries.
Want a route through all this instead of an archive? The track
puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.
Source: Omni-MATH,
licensed Apache-2.0.
Statement reproduced verbatim; metadata (topic, difficulty) added by this project.