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1520052053147 is a prime number
BaseRepresentation
bin10110000111101010001…
…010011010010010011011
312101022112001002222110122
4112013222022122102123
5144401031311200042
63122145142503455
7214551210160625
oct26075212322233
95338461088418
101520052053147
11536717835741
1220671a434b8b
13b0456818004
14537dc821c15
15298179987d2
hex161ea29a49b

1520052053147 has 2 divisors, whose sum is σ = 1520052053148. Its totient is φ = 1520052053146.

The previous prime is 1520052053111. The next prime is 1520052053167. The reversal of 1520052053147 is 7413502500251.

It is a strong prime.

It is an emirp because it is prime and its reverse (7413502500251) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-1520052053147 is a prime.

It is a super-2 number, since 2×15200520531472 (a number of 25 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (1520052053167) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 760026026573 + 760026026574.

It is an arithmetic number, because the mean of its divisors is an integer number (760026026574).

Almost surely, 21520052053147 is an apocalyptic number.

1520052053147 is a deficient number, since it is larger than the sum of its proper divisors (1).

1520052053147 is an equidigital number, since it uses as much as digits as its factorization.

1520052053147 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 42000, while the sum is 35.

Adding to 1520052053147 its reverse (7413502500251), we get a palindrome (8933554553398).

The spelling of 1520052053147 in words is "one trillion, five hundred twenty billion, fifty-two million, fifty-three thousand, one hundred forty-seven".