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151220143 is a prime number
BaseRepresentation
bin10010000001101…
…10111110101111
3101112112210001001
421000312332233
5302203021033
623001101131
73514231024
oct1100667657
9345483031
10151220143
11783a5a28
12427877a7
1325438399
141612554b
15d420e7d
hex9036faf

151220143 has 2 divisors, whose sum is σ = 151220144. Its totient is φ = 151220142.

The previous prime is 151220137. The next prime is 151220149. The reversal of 151220143 is 341022151.

It is a balanced prime because it is at equal distance from previous prime (151220137) and next prime (151220149).

It is an emirp because it is prime and its reverse (341022151) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 151220143 - 29 = 151219631 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 151220143.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (151220149) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 75610071 + 75610072.

It is an arithmetic number, because the mean of its divisors is an integer number (75610072).

Almost surely, 2151220143 is an apocalyptic number.

151220143 is a deficient number, since it is larger than the sum of its proper divisors (1).

151220143 is an equidigital number, since it uses as much as digits as its factorization.

151220143 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 240, while the sum is 19.

The square root of 151220143 is about 12297.1599566729. The cubic root of 151220143 is about 532.7660574690.

Adding to 151220143 its reverse (341022151), we get a palindrome (492242294).

The spelling of 151220143 in words is "one hundred fifty-one million, two hundred twenty thousand, one hundred forty-three".