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5302540553537 is a prime number
BaseRepresentation
bin100110100101001100000…
…0010111100110101000001
3200202220210011112122102202
41031022120002330311001
51143334100340203122
615135541542322545
71055044620015233
oct115123002746501
920686704478382
105302540553537
1117648819a6287
12717802129455
132c6048350a96
14144904094253
1592de7e6be92
hex4d2980bcd41

5302540553537 has 2 divisors, whose sum is σ = 5302540553538. Its totient is φ = 5302540553536.

The previous prime is 5302540553491. The next prime is 5302540553591. The reversal of 5302540553537 is 7353550452035.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 4073034366976 + 1229506186561 = 2018176^2 + 1108831^2 .

It is a cyclic number.

It is not a de Polignac number, because 5302540553537 - 210 = 5302540552513 is a prime.

It is a super-4 number, since 4×53025405535374 (a number of 52 digits) contains 4444 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 5302540553491 and 5302540553500.

It is not a weakly prime, because it can be changed into another prime (5302540553437) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2651270276768 + 2651270276769.

It is an arithmetic number, because the mean of its divisors is an integer number (2651270276769).

Almost surely, 25302540553537 is an apocalyptic number.

It is an amenable number.

5302540553537 is a deficient number, since it is larger than the sum of its proper divisors (1).

5302540553537 is an equidigital number, since it uses as much as digits as its factorization.

5302540553537 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 4725000, while the sum is 47.

The spelling of 5302540553537 in words is "five trillion, three hundred two billion, five hundred forty million, five hundred fifty-three thousand, five hundred thirty-seven".