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133534045041150 = 2352132171097282461
BaseRepresentation
bin11110010111001011010001…
…010100011111000111111110
3122111210201201012020120010110
4132113023101110133013332
5120000310211012304100
61152000405440050450
740061342056003035
oct3627132124370776
9574721635216113
10133534045041150
113960351a222746
1212b87961603a26
13596828b659900
1424d9121a3d71c
1510687d871a250
hex7972d151f1fe

133534045041150 has 288 divisors, whose sum is σ = 380039645825568. Its totient is φ = 30908083814400.

The previous prime is 133534045041149. The next prime is 133534045041223. The reversal of 133534045041150 is 51140540435331.

It is a Harshad number since it is a multiple of its sum of digits (39).

It is a junction number, because it is equal to n+sod(n) for n = 133534045041099 and 133534045041108.

It is a congruent number.

It is an unprimeable number.

It is a polite number, since it can be written in 143 ways as a sum of consecutive naturals, for example, 472610920 + ... + 472893380.

It is an arithmetic number, because the mean of its divisors is an integer number (1319582103561).

Almost surely, 2133534045041150 is an apocalyptic number.

133534045041150 is a gapful number since it is divisible by the number (10) formed by its first and last digit.

It is a practical number, because each smaller number is the sum of distinct divisors of 133534045041150, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (190019822912784).

133534045041150 is an abundant number, since it is smaller than the sum of its proper divisors (246505600784418).

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

133534045041150 is a wasteful number, since it uses less digits than its factorization.

133534045041150 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 283616 (or 283598 counting only the distinct ones).

The product of its (nonzero) digits is 216000, while the sum is 39.

Adding to 133534045041150 its reverse (51140540435331), we get a palindrome (184674585476481).

The spelling of 133534045041150 in words is "one hundred thirty-three trillion, five hundred thirty-four billion, forty-five million, forty-one thousand, one hundred fifty".