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14253242533151 is a prime number
BaseRepresentation
bin1100111101101001011101…
…1101010100010100011111
31212110121002222102120011012
43033122113131110110133
53332011120042030101
650151503233245435
73000522431355536
oct317322735242437
955417088376135
1014253242533151
1145a5847912aa6
12172245a10387b
137c50c8a241bb
14373c0977b31d
1519ab5d717bbb
hexcf69775451f

14253242533151 has 2 divisors, whose sum is σ = 14253242533152. Its totient is φ = 14253242533150.

The previous prime is 14253242533121. The next prime is 14253242533181. The reversal of 14253242533151 is 15133524235241.

It is a balanced prime because it is at equal distance from previous prime (14253242533121) and next prime (14253242533181).

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-14253242533151 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 14253242533099 and 14253242533108.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (14253242533121) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7126621266575 + 7126621266576.

It is an arithmetic number, because the mean of its divisors is an integer number (7126621266576).

Almost surely, 214253242533151 is an apocalyptic number.

14253242533151 is a deficient number, since it is larger than the sum of its proper divisors (1).

14253242533151 is an equidigital number, since it uses as much as digits as its factorization.

14253242533151 is an evil number, because the sum of its binary digits is even.

The product of its digits is 432000, while the sum is 41.

Adding to 14253242533151 its reverse (15133524235241), we get a palindrome (29386766768392).

The spelling of 14253242533151 in words is "fourteen trillion, two hundred fifty-three billion, two hundred forty-two million, five hundred thirty-three thousand, one hundred fifty-one".