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1305340152434 = 219257133661699
BaseRepresentation
bin10010111111101100010…
…101100000101001110010
311121210022101210022012122
4102333230111200221302
5132341314034334214
62435355345354242
7163210355140043
oct22775425405162
94553271708178
101305340152434
1146365653880a
12190b97938982
1396129571b56
14472708b28ca
1523e4cb6c88e
hex12fec560a72

1305340152434 has 16 divisors (see below), whose sum is σ = 2069083116000. Its totient is φ = 615913104384.

The previous prime is 1305340152427. The next prime is 1305340152437. The reversal of 1305340152434 is 4342510435031.

It is a super-2 number, since 2×13053401524342 (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1305340152394 and 1305340152403.

It is not an unprimeable number, because it can be changed into a prime (1305340152437) by changing a digit.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 66821084 + ... + 66840615.

It is an arithmetic number, because the mean of its divisors is an integer number (129317694750).

Almost surely, 21305340152434 is an apocalyptic number.

1305340152434 is a deficient number, since it is larger than the sum of its proper divisors (763742963566).

1305340152434 is a wasteful number, since it uses less digits than its factorization.

1305340152434 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 133661977.

The product of its (nonzero) digits is 86400, while the sum is 35.

Adding to 1305340152434 its reverse (4342510435031), we get a palindrome (5647850587465).

The spelling of 1305340152434 in words is "one trillion, three hundred five billion, three hundred forty million, one hundred fifty-two thousand, four hundred thirty-four".

Divisors: 1 2 19 38 257 514 4883 9766 133661699 267323398 2539572281 5079144562 34351056643 68702113286 652670076217 1305340152434