561

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Summary

Factorization

561 is a square-free number with prime factors 3, 11, and 17. The prime factorization is:

$\!561=3*11*17$

Properties and families

Property or family	Parameter values	First few numbers satisfying the property	Proof of satisfaction/membership/containment
Carmichael number (also called absolute pseudoprime)	first Carmichael number	561, 1105, 1729, 2465, 2821, 6601, [SHOW MORE] 8911, 10585, 15841, 29341, 41041, 46657, 52633, 62745, 63973, 75361, 101101, 115921, 126217, 162401, 172081, 188461, 252601, 278545, 294409, 314821, 334153, 340561, 399001, 410041, 449065, 488881, 512461 View list on OEIS	The universal exponent is $\operatorname {lcm} \{3-1,11-1,17-1\}=80$ which divides $561-1=560$
Poulet number (Fermat pseudoprime to base 2)	second Poulet number	341, 561, 645, 1105, 1387, 1729, 1905, 2047, [SHOW MORE] 2465, 2701, 2821, 3277, 4033, 4369, 4371, 4681, 5461, 6601, 7957, 8321, 8481, 8911, 10261, 10585, 11305, 12801, 13741, 13747, 13981, 14491, 15709, 15841, 16705, 18705, 18721, 19951, 23001, 23377, 25761, 29341 View list on OEIS	follows from its being an odd Carmichael number.

Arithmetic functions

Function	Value	Explanation
Euler totient function	320	The Euler totient function is $(3-1)(11-1)(17-1)=(2)(10)(16)=320$.
universal exponent	80	The universal exponent is the least common multiple of $\{3-1,11-1,17-1\}$, which is $\operatorname {lcm} \{2,10,16\}$ and equals 80. Note that 561 is a Carmichael number precisely because the universal exponent divides 561 - 1 = 560.
Mobius function	-1	The number is a square-free number and it has an odd number of prime divisors (3 prime divisors).
divisor count function	8	$(1+1)(1+1)(1+1)$ where the first 1s in each sum denote the multiplicities of the prime divisors.
largest prime divisor	17
largest prime power divisor	17

Structure of integers mod 561

Abstract structure

By the Chinese remainder theorem, we have the ring decomposition:

$\mathbb {Z} /561\mathbb {Z} \cong \mathbb {Z} /3\mathbb {Z} \times \mathbb {Z} /11\mathbb {Z} \times \mathbb {Z} /17\mathbb {Z}$

For the multiplicative group, this gives:

$(\mathbb {Z} /561\mathbb {Z} )^{*}\cong (\mathbb {Z} /3\mathbb {Z} )^{*}\times (\mathbb {Z} /11\mathbb {Z} )^{*}\times (\mathbb {Z} /17\mathbb {Z} )^{*}$

All groups on the right side are cyclic, and we get:

$(\mathbb {Z} /561\mathbb {Z} )^{*}\cong \mathbb {Z} /2\mathbb {Z} \times \mathbb {Z} /10\mathbb {Z} \times \mathbb {Z} /16\mathbb {Z}$

This group has order $2\times 10\times 16=320$ (the order of the multiplicative group is termed the Euler totient function and denoted $\varphi$, so this is saying that $\varphi (561)=320$).

The exponents of the factor groups are 2, 10, and 16 respectively, hence the exponent of the whole group is the lcm of these, which is 80. The exponent of the multiplicative group is termed the universal exponent and denoted $\lambda$, so this is saying that $\lambda (561)=80$).