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151300117 is a prime number
BaseRepresentation
bin10010000010010…
…10100000010101
3101112200211202001
421001022200111
5302213100432
623002515301
73515013133
oct1101124025
9345624661
10151300117
1178450021
1242805b31
13254658c7
1416146753
15d4399e7
hex904a815

151300117 has 2 divisors, whose sum is σ = 151300118. Its totient is φ = 151300116.

The previous prime is 151300081. The next prime is 151300133. The reversal of 151300117 is 711003151.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 126360081 + 24940036 = 11241^2 + 4994^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-151300117 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 151300094 and 151300103.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (151300147) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 75650058 + 75650059.

It is an arithmetic number, because the mean of its divisors is an integer number (75650059).

Almost surely, 2151300117 is an apocalyptic number.

It is an amenable number.

151300117 is a deficient number, since it is larger than the sum of its proper divisors (1).

151300117 is an equidigital number, since it uses as much as digits as its factorization.

151300117 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 105, while the sum is 19.

The square root of 151300117 is about 12300.4112532874. The cubic root of 151300117 is about 532.8599601351.

Adding to 151300117 its reverse (711003151), we get a palindrome (862303268).

The spelling of 151300117 in words is "one hundred fifty-one million, three hundred thousand, one hundred seventeen".