Bill Gosper: "Yeah, what about 2*10^4679+3, (two zillion and three)-?" Phil Carmody: "It seems to precede 2*10^4772+2003 (two zillion two thousand and three)" It's instructive here to look at the original (without using 'zillion') answer: 2*10^63 + 2*10^36 + 2*10^12 + 2*10^3 + 2*10^2 + 93 (two vigintillion, two undecillion, two trillion, two thousand, two hundred, ninety, three). The 93 is forced because it is the only two- digit number that will make the entire quantity prime, but 23 (twenty, three) would have been a more desirable choice to force the written number to the back of the dictionary. Just 2 (two) would have been even better but a number ending in 2 cannot be prime. So, let's prepend the 'two zillion' (2*10^n) and force the presumably- optimal 23 ending. 2*10^n (two zillion) +2*10^63 (two vigintillion) +2*10^36 (two undecillion) +2*10^12 (two trillion) +2000 (two thousand) +200 (two hundred) +20 (twenty) +3 (three) Of course, n >=66 because we have perfectly good, widely-dictionaried names for numbers less than this. A quick run has n = {77, 113, 116, 158, 342, 464, 468, 565, 2171, 2274, 2340, 3347, 5724, ...}. That might have been the end of it, but I noticed that there is nothing to prevent our indefinite 'zillion' from being repeated. And every repetition forces the written number further back in the dictionary. So one begins to look at the divisors of n to see which combination results in the most number of greater-than-10^66 (zillion) amounts. For example, 10^5724 = 10^1431 * 10^1431 * 10^1431 * 10^1431, so, taking a zillion to be 10^1431, our large prime starts out with "two zillion zillion zillion zillion, two vigintillion, ..." Our n = 2340 solution allows for 30 written zillions (of 10^78 each).