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49149712384753 is a prime number
BaseRepresentation
bin10110010110011100011110…
…01001011100001011110001
320110000122222221222222120211
423023032033021130023301
522420232102332303003
6252311022142504121
713231643456064661
oct1313161711341361
9213018887888524
1049149712384753
111472a3043467a2
125619663295041
132156a4b253c66
14c1cc046841a1
155a376bb1396d
hex2cb38f25c2f1

49149712384753 has 2 divisors, whose sum is σ = 49149712384754. Its totient is φ = 49149712384752.

The previous prime is 49149712384579. The next prime is 49149712384781. The reversal of 49149712384753 is 35748321794194.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 30905683066944 + 18244029317809 = 5559288^2 + 4271303^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-49149712384753 is a prime.

It is not a weakly prime, because it can be changed into another prime (49149712382753) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 24574856192376 + 24574856192377.

It is an arithmetic number, because the mean of its divisors is an integer number (24574856192377).

Almost surely, 249149712384753 is an apocalyptic number.

It is an amenable number.

49149712384753 is a deficient number, since it is larger than the sum of its proper divisors (1).

49149712384753 is an equidigital number, since it uses as much as digits as its factorization.

49149712384753 is an evil number, because the sum of its binary digits is even.

The product of its digits is 182891520, while the sum is 67.

The spelling of 49149712384753 in words is "forty-nine trillion, one hundred forty-nine billion, seven hundred twelve million, three hundred eighty-four thousand, seven hundred fifty-three".