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55015004003 is a prime number
BaseRepresentation
bin110011001111001001…
…100101011101100011
312021000002102202210212
4303033021211131203
51400132320112003
641135030244335
73655215550451
oct631711453543
9167002382725
1055015004003
1121371598863
12a7b44b66ab
135259a52a0b
14293c7b99d1
15166ecb1dd8
hexccf265763

55015004003 has 2 divisors, whose sum is σ = 55015004004. Its totient is φ = 55015004002.

The previous prime is 55015003993. The next prime is 55015004041. The reversal of 55015004003 is 30040051055.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 55015004003 - 234 = 37835134819 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 55015004003.

It is not a weakly prime, because it can be changed into another prime (55015000003) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 27507502001 + 27507502002.

It is an arithmetic number, because the mean of its divisors is an integer number (27507502002).

Almost surely, 255015004003 is an apocalyptic number.

55015004003 is a deficient number, since it is larger than the sum of its proper divisors (1).

55015004003 is an equidigital number, since it uses as much as digits as its factorization.

55015004003 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1500, while the sum is 23.

Adding to 55015004003 its reverse (30040051055), we get a palindrome (85055055058).

The spelling of 55015004003 in words is "fifty-five billion, fifteen million, four thousand, three".