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40132351451 is a prime number
BaseRepresentation
bin100101011000000100…
…110001010111011011
310211120220002200120002
4211120010301113123
51124142340221301
630234143323215
72620342234043
oct453004612733
9124526080502
1040132351451
1116024736705
1279402b050b
133a2761242a
141d29dc0723
15109d42c16b
hex9581315db

40132351451 has 2 divisors, whose sum is σ = 40132351452. Its totient is φ = 40132351450.

The previous prime is 40132351351. The next prime is 40132351453. The reversal of 40132351451 is 15415323104.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 40132351451 - 230 = 39058609627 is a prime.

It is a super-2 number, since 2×401323514512 (a number of 22 digits) contains 22 as substring.

Together with 40132351453, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (40132351453) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 20066175725 + 20066175726.

It is an arithmetic number, because the mean of its divisors is an integer number (20066175726).

Almost surely, 240132351451 is an apocalyptic number.

40132351451 is a deficient number, since it is larger than the sum of its proper divisors (1).

40132351451 is an equidigital number, since it uses as much as digits as its factorization.

40132351451 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 7200, while the sum is 29.

Adding to 40132351451 its reverse (15415323104), we get a palindrome (55547674555).

The spelling of 40132351451 in words is "forty billion, one hundred thirty-two million, three hundred fifty-one thousand, four hundred fifty-one".