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152461100130047 is a prime number
BaseRepresentation
bin100010101010100110011101…
…101101110001111011111111
3201222211010202012120012211222
4202222212131231301323333
5124440410313113130142
61300131355033103555
744053644121534412
oct4252463555617377
9658733665505758
10152461100130047
1144640431409495
1215123b95b57bbb
13670c035093707
142991229a95b79
151295ce2e17bd2
hex8aa99db71eff

152461100130047 has 2 divisors, whose sum is σ = 152461100130048. Its totient is φ = 152461100130046.

The previous prime is 152461100130019. The next prime is 152461100130083. The reversal of 152461100130047 is 740031001164251.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 152461100130047 - 28 = 152461100129791 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 152461100129989 and 152461100130016.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (152461100130247) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 76230550065023 + 76230550065024.

It is an arithmetic number, because the mean of its divisors is an integer number (76230550065024).

Almost surely, 2152461100130047 is an apocalyptic number.

152461100130047 is a deficient number, since it is larger than the sum of its proper divisors (1).

152461100130047 is an equidigital number, since it uses as much as digits as its factorization.

152461100130047 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 20160, while the sum is 35.

Adding to 152461100130047 its reverse (740031001164251), we get a palindrome (892492101294298).

The spelling of 152461100130047 in words is "one hundred fifty-two trillion, four hundred sixty-one billion, one hundred million, one hundred thirty thousand, forty-seven".