Those Fascinating Numbers 33
99
the numerical representation adopted by the Greeks to denote the word AMEN
= αµην = 1 + 40 + 8 + 50 = 99 (see Ore [156], p. 28);
the third solution x of the Fermat-Pell equation x2 2y2 = 1: it is well known
that this equation has infinitely many solutions, the first nine being (x, y) =
(3, 2), (17,12), (99,70), (577,408), (3363,2378), (19601,13860), (114243,80782),
(665857,470832) and (3880899,2744210).
100
the only solution n
1012
of σ(n) = 2n + 17 (see the number 196);
the largest known solution y of the problem consisting in finding a right an-
gle triangle whose sides x, y, z (all integers) represent respectively a triangu-
lar number, a perfect square and a pentagonal number: here the solution is
(x, y, z) = (105, 100, 145); the only other solution is (3, 4, 5) (R.K. Guy [101],
D21).
101
the largest known prime number of the form 10n + 1; any other prime number,
with more than three digits and of the form 10n + 1 must be such that n
131071 (see the number 19 841 for the factorization of those numbers of the
form
102r
+ 1 for 1 r 12);
the only prime
number39
whose digits are 1 and 0 in alternation;
the first member p of the second quintuple (p, p + 2, p + 6, p + 8, p + 12) made
up entirely of prime numbers: the smallest quintuple satisfying this property
is (11, 13, 17, 19, 23);
the third and possibly the largest prime number whose sum of digits is equal
to 2, the other two known such numbers being 2 and 11 (see the number 389);
the smallest odd prime number p such that p k! is composite for 1 k! p
(an observation due to Felice Russo).
39The
proof of this result is easy. Indeed, given a number of the form 101010. . . 1 containing k
times the digit 1, if we multiply it by 11, we obtain the number 111. . . 1 containing 2k times the
digit 1, in which case
101010 . . . 1
k
times the digit 1
×11 = 111 . . . 1
2k
= 111 . . . 1
k
×100 . . . 01,
where this last number contains k 1 times the digit 0. Thus, by multiplying the original number
by 11, we obtain the product of two numbers larger than 11, which proves that the original number
cannot be prime.
Previous Page Next Page