A beastly number (A051003) is a number with substring 666.
It is easy enough to prove that for each positive integer n, there is an integer k such that kn is beastly. Let f(n) be the smallest such k.
Show that f(n) is bounded over all integers n. What is the largest possible value of f(n)?
the largest possible value is 4263 , which occurs as f(15625) , and occurs infinitely often as f(15625 * (10^t + 1)) for sufficiently large t . as allan's post suggests, it is easy to get a universal upper bound on f(n) by considering initial digits. indeed, let m = ceiling(log_10(n)) so that 10^(m-1) < n <= 10^m , and let k = ceiling(666 * 10^m / n) . then k <= ceiling(666 * 10^m / (10^(m-1) + 1)) <= 6660 , and kn begins with the digits 666 , so that f(n) <= 6660 , independently of n . to get the largest value of f(n) that occurs, i found it easier to consider final digits and split it into numerous cases; perhaps there is an easier calculation using initial digits. note that f(n) = f(10n) , so we need only consider those n not divisible by 10 . if gcd(n, 10) = 1 , then kn = 666 mod 1000 for some 0 < k < 1000 , so f(n) < 1000 . if n is even, consider the largest power of 2 that divides n . as noted, we may assume that 5 does not divide n . if the largest power of 2 that divides n is 2^1 , then kn = 666 mod 1000 for some 0 < k < 500 , whence f(n) < 500 . if the largest power of 2 that divides n is 2^2 , then kn = 6664 mod 10000 for some 0 < k < 2500 , whence f(n) < 2500 . if the largest power of 2 that divides n is 2^3 , then kn = 6664 mod 10000 for some 0 < k < 1250 , whence f(n) < 1250 . if the largest power of 2 that divides n is 2^4 , then kn = 66608 mod 100000 for some 0 < k < 6250 , whence f(n) < 6250 . if n is divisibly by 2^5 , then kn = 66624 mod 100000 for some 0 < k < 3125 , whence f(n) < 3125 . now consider the case when n is divisible by 5 , but not by 2 . if the largest power of 5 that divides n is 5^1 , then kn = 6665 mod 10000 for some 0 < k < 2000 , whence f(n) < 2000 . if the largest power of 5 that divides n is 5^2 , then kn = 66625 mod 100000 for some 0 < k < 4000 , whence f(n) < 4000 . if the largest power of 5 that divides n is 5^3 , then kn = 66625 mod 100000 for some 0 < k < 800 , whence f(n) < 800 . if the largest power of 5 that divides n is 5^4 , then kn = 666875 mod 1000000 for some 0 < k < 1600 , whence f(n) < 1600 . if the largest power of 5 that divides n is 5^5 , then kn = 6665625 mod 10000000 for some 0 < k < 3200 , whence f(n) < 3200 . if the largest power of 5 that divides n is 5^6 , then kn = 66609375 mod 100000000 for some 0 < k < 6400 , whence f(n) < 6400 . if the largest power of 5 that divides n is 5^7 , then kn = 66640625 mod 100000000 for some 0 < k < 1280 , whence f(n) < 1280 . if the largest power of 5 that divides n is 5^8 , then kn = 666015625 mod 1000000000 for some 0 < k < 2560 , whence f(n) < 2560 . if n is divisible by 5^9 , then kn = 666015625 mod 1000000000 for some 0 < k < 512 , whence f(n) < 512 . these give the slightly better upper bound f(n) <= 6249 . however, we can consider each of these cases in more detail, as i had my computer do. for example, suppose n is divisible by 5^8 , but not by 5^9 , nor by 2 . there are 1024 residue classes of such n modulo 10^9 , and for each, we verify that there is 0 < k <= 1705 such that the digit string 666 occurs within the last 9 digits of kn . (this only requires that kn is one of the residues: 666015625, 666406250 or 666796875 .) this shows that f(n) <= 1705 for all n in that case above. i checked all the cases above, and found that f(n) <= 4263 in all cases, with smaller upper bounds in all cases except for the case when 5^6 and 2^0 are the largest powers of 5 and 2 dividing n . (it took much longer to code it than it took to execute (1.1 secs), and there is some possibility of error.) moreover, 4263 indeed occurs as f(5^6) , so that is the largest value. i did not give it much thought, but i suppose it is conceivable that the upper bound from examining the cases in detail could not be obtained. if that happens, perhaps it would suffice to extend the cases to consider higher powers of 2 and/or 5 . mike