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132001231020120 = 233257211680278453
BaseRepresentation
bin11110000000110111101110…
…011111000010000001011000
3122022101012020202020102212200
4132000313232133002001120
5114300202010120120440
61144424310044010200
736542526362542500
oct3600675637020130
9568335222212780
10132001231020120
1139072453711710
121297a881230960
13588687cb82117
142484c72334a00
15103d9c568cc30
hex780dee7c2058

132001231020120 has 288 divisors, whose sum is σ = 544413241167120. Its totient is φ = 27428827184640.

The previous prime is 132001231020079. The next prime is 132001231020121. The reversal of 132001231020120 is 21020132100231.

132001231020120 is a `hidden beast` number, since 1 + 3 + 200 + 12 + 310 + 20 + 120 = 666.

It is a super-4 number, since 4×1320012310201204 (a number of 58 digits) contains 4444 as substring.

It is a Harshad number since it is a multiple of its sum of digits (18).

It is a junction number, because it is equal to n+sod(n) for n = 132001231020093 and 132001231020102.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (132001231020121) by changing a digit.

It is a polite number, since it can be written in 71 ways as a sum of consecutive naturals, for example, 339945187 + ... + 340333266.

Almost surely, 2132001231020120 is an apocalyptic number.

132001231020120 is a gapful number since it is divisible by the number (10) formed by its first and last digit.

It is an amenable number.

132001231020120 is an abundant number, since it is smaller than the sum of its proper divisors (412412010147000).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

132001231020120 is a wasteful number, since it uses less digits than its factorization.

132001231020120 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 680278495 (or 680278481 counting only the distinct ones).

The product of its (nonzero) digits is 144, while the sum is 18.

Adding to 132001231020120 its reverse (21020132100231), we get a palindrome (153021363120351).

The spelling of 132001231020120 in words is "one hundred thirty-two trillion, one billion, two hundred thirty-one million, twenty thousand, one hundred twenty".