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1352422051433 is a prime number
BaseRepresentation
bin10011101011100010101…
…000101100111001101001
311210021211121200120210122
4103223202220230321221
5134124230021121213
62513143310053025
7166465164522515
oct23534250547151
94707747616718
101352422051433
11481616a99342
1219a1374aa175
139a6c08a3b9b
1449659839145
15252a6249d08
hex13ae2a2ce69

1352422051433 has 2 divisors, whose sum is σ = 1352422051434. Its totient is φ = 1352422051432.

The previous prime is 1352422051429. The next prime is 1352422051439. The reversal of 1352422051433 is 3341502242531.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 1002535605289 + 349886446144 = 1001267^2 + 591512^2 .

It is a cyclic number.

It is not a de Polignac number, because 1352422051433 - 22 = 1352422051429 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 1352422051393 and 1352422051402.

It is not a weakly prime, because it can be changed into another prime (1352422051439) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 676211025716 + 676211025717.

It is an arithmetic number, because the mean of its divisors is an integer number (676211025717).

Almost surely, 21352422051433 is an apocalyptic number.

It is an amenable number.

1352422051433 is a deficient number, since it is larger than the sum of its proper divisors (1).

1352422051433 is an equidigital number, since it uses as much as digits as its factorization.

1352422051433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 86400, while the sum is 35.

Adding to 1352422051433 its reverse (3341502242531), we get a palindrome (4693924293964).

The spelling of 1352422051433 in words is "one trillion, three hundred fifty-two billion, four hundred twenty-two million, fifty-one thousand, four hundred thirty-three".