The sequence you provided is known as the powers of 11. Each term represents the coefficients of the binomial expansion of ((a + b)^n) where (n) is the term's index. The next terms in the sequence would be 161051 (for (n=5)) and 1771561 (for (n=6)). Thus, the complete sequence is 1, 11, 121, 1331, 14641, 161051, 1771561.
1, 2, 11, 22, 121, 242, 1331, 2662.
The factors of 121 are 1•121 and 11•11 :)
121 121 121 121
1, 2, 3, 5, 6, 10, 15, 30, 41, 82, 123, 205, 246, 410, 615, 1230.
1, 2, 3, 5, 6, 10, 11, 15, 22, 30, 33, 55, 66, 110, 121, 165, 242, 330, 363, 605, 726, 1210, 1331, 1815, 2662, 3630, 3993, 6655, 7986, 13310, 19965, 39930.
11, 11^2=121, 11^3=1331, 11^4=14641 etc
14641
121 x 121
1, 11, 121, 1331.
1212 = 14,641 The factors of 14,641 are 1, 11, 121, 1331, and 14641. The prime factors of 14,641 are 11, 11, 11, and 11. The prime factorization of 14,641 is 11 x 11 x 11 x 11 = 114
#include#includevoid main(){clrscr();for (int i=1;i
It can be: 11 times 121 = 1331.
1, 11, 121, 1331.
121
eleven 11 x 11 = 121 121 x 11 = 1331
6 over 11 Another answer: 1296 over 14641 = (36 over 121)2
It equals 14641.