Base | Representation |
---|---|
bin | 100010111000010001001100… |
… | …111001001110011000000111 |
3 | 202010010221002120212112010012 |
4 | 202320101030321032120013 |
5 | 130101302400023412242 |
6 | 1302131051054220435 |
7 | 44211544323033314 |
oct | 4270211471163007 |
9 | 663127076775105 |
10 | 153400342013447 |
11 | 449727a0a23081 |
12 | 152560201b471b |
13 | 6779789825069 |
14 | 29c488aaa3d0b |
15 | 12b04655c9282 |
hex | 8b844ce4e607 |
153400342013447 has 2 divisors, whose sum is σ = 153400342013448. Its totient is φ = 153400342013446.
The previous prime is 153400342013441. The next prime is 153400342013543. The reversal of 153400342013447 is 744310243004351.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 153400342013447 - 24 = 153400342013431 is a prime.
It is a super-3 number, since 3×1534003420134473 (a number of 44 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (153400342013441) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 76700171006723 + 76700171006724.
It is an arithmetic number, because the mean of its divisors is an integer number (76700171006724).
Almost surely, 2153400342013447 is an apocalyptic number.
153400342013447 is a deficient number, since it is larger than the sum of its proper divisors (1).
153400342013447 is an equidigital number, since it uses as much as digits as its factorization.
153400342013447 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 483840, while the sum is 41.
Adding to 153400342013447 its reverse (744310243004351), we get a palindrome (897710585017798).
The spelling of 153400342013447 in words is "one hundred fifty-three trillion, four hundred billion, three hundred forty-two million, thirteen thousand, four hundred forty-seven".
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