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39970479313 is a prime number
BaseRepresentation
bin100101001110011011…
…010001110011010001
310211011121111201201201
4211032123101303101
51123324420314223
630210122031201
72613335316442
oct451633216321
9124147451651
1039970479313
1115a51325786
1278b6048501
1339ccc177bb
141d126c52c9
15108e104ead
hex94e6d1cd1

39970479313 has 2 divisors, whose sum is σ = 39970479314. Its totient is φ = 39970479312.

The previous prime is 39970479277. The next prime is 39970479317. The reversal of 39970479313 is 31397407993.

39970479313 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 36790308864 + 3180170449 = 191808^2 + 56393^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-39970479313 is a prime.

It is not a weakly prime, because it can be changed into another prime (39970479317) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 19985239656 + 19985239657.

It is an arithmetic number, because the mean of its divisors is an integer number (19985239657).

Almost surely, 239970479313 is an apocalyptic number.

It is an amenable number.

39970479313 is a deficient number, since it is larger than the sum of its proper divisors (1).

39970479313 is an equidigital number, since it uses as much as digits as its factorization.

39970479313 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 3857868, while the sum is 55.

The spelling of 39970479313 in words is "thirty-nine billion, nine hundred seventy million, four hundred seventy-nine thousand, three hundred thirteen".