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Using the Euclidean algorithm: 126 = 1·84 + 42, then 84 = 2·42 + 0, so the last nonzero remainder is 42. Equivalently, 84 = 2²·3·7 and 126 = 2·3²·7, and the gcd takes the minimum exponent of each prime, giving 2·3·7 = 42.
A number is prime when it has no divisor other than 1 and itself; it suffices to check primes up to its square root. Since √97 ≈ 9.8, we test 2, 3, 5 and 7, and none divides 97, so 97 is prime. The other three numbers each factor into smaller factors greater than 1.
Squaring the congruence gives n² ≡ 3² = 9 (mod 7). Reducing 9 modulo 7 leaves 2, which is already the least nonnegative residue. This uses the rule that congruences may be multiplied term by term.
Compute 7⁴ = 2401 directly, whose units digit is 1, so the remainder on division by 10 is 1. Equivalently, work modulo 10: 7² ≡ 49 ≡ 9, then 7⁴ ≡ 9² = 81 ≡ 1 (mod 10).
Bézout's identity states that the set of integer combinations ax + by equals the set of multiples of gcd(a, b). Hence c is representable exactly when gcd(a, b) divides c, and when a solution exists all solutions follow from one particular solution.
If n = p₁^a₁···p_k^a_k, the number of positive divisors is the product (a₁+1)···(a_k+1). Here this is (3+1)(2+1)(1+1) = 4·3·2 = 24. Each divisor is determined uniquely by choosing an exponent from 0 to a_i for each prime.
φ(n) counts the integers from 1 to n that are coprime to n. Among 1 through 12, the numbers coprime to 12 are 1, 5, 7 and 11, so φ(12) = 4. The formula φ(12) = 12·(1−1/2)·(1−1/3) gives the same result.
Since gcd(3, 7) = 1, the congruence has a unique solution modulo 7. Testing values, 3·5 = 15 ≡ 1 (mod 7), so x ≡ 5 works; equivalently multiply both sides by the inverse 5 of 3 modulo 7 to get x ≡ 5.
Fermat's little theorem states that for prime p and integer a not divisible by p, a^(p−1) ≡ 1 (mod p). A standard proof multiplies the nonzero residues 1, 2, …, p−1 by a and observes that the resulting set is a permutation of the same residues.
Since 3 is not divisible by 7, Fermat's little theorem gives 3⁶ ≡ 1 (mod 7). Writing 200 = 6·33 + 2, we get 3²⁰⁰ ≡ (3⁶)³³·3² ≡ 3² = 9 ≡ 2, so the remainder is 2.
The congruence ax ≡ b (mod n) is solvable exactly when gcd(a, n) divides b. For 4x ≡ 3 (mod 10), gcd(4, 10) = 2 does not divide 3, so no solution exists. In the other three cases the gcd divides the right-hand side.
The Chinese remainder theorem states that the system is solvable for every choice of a and b, with a unique solution modulo mn, exactly when the moduli are coprime, that is gcd(m, n) = 1. Coprimality lets us combine the two conditions independently.
The numbers congruent to 3 modulo 5 are 3, 8, 13, 18, …; checking modulo 3, we find 8 ≡ 2 (mod 3). Since gcd(3, 5) = 1, the solution is unique modulo 15, so 8 is the smallest positive solution.
Since 45 = 3²·5, the totient is φ(45) = 45·(1−1/3)·(1−1/5) = 45·(2/3)·(4/5) = 24. Alternatively, φ(3²) = 6 and φ(5) = 4, and coprimality gives φ(45) = 6·4 = 24.
Fermat's little theorem gives 2¹² ≡ 1 (mod 13) because 13 is prime and 2 is not divisible by 13. Since 100 = 12·8 + 4, we get 2¹⁰⁰ ≡ 2⁴ = 16 ≡ 3 (mod 13).
Apply the algorithm: 1071 = 2·462 + 147, 462 = 3·147 + 21, 147 = 7·21 + 0. The last nonzero remainder is 21, so gcd(1071, 462) = 21. Checking the factorization, 1071 = 3·7·51 and 462 = 2·3·7·11, sharing 3·7.
Because 11 is prime and 5 is not divisible by 11, Fermat's little theorem gives 5¹⁰ ≡ 1 (mod 11). Since 100 is a multiple of 10, 5¹⁰⁰ = (5¹⁰)¹⁰ ≡ 1¹⁰ ≡ 1 (mod 11).
For 12 = 2²·3, the divisor count is (2+1)(1+1) = 6. The divisors are 1, 2, 3, 4, 6 and 12, which confirms the count.
We need 8x ≡ 1 (mod 13). Testing x = 5 gives 8·5 = 40 = 3·13 + 1 ≡ 1, so 5 is the inverse. Since gcd(8, 13) = 1 the inverse exists and is unique modulo 13.
The contrapositive of 'P implies Q' is 'not Q implies not P'. Here P is 'n² is even' and Q is 'n is even', so the contrapositive is 'if n is not even, then n² is not even', that is, 'if n² is odd then n is odd'. The contrapositive is logically equivalent to the original claim.
Checking residues 0 to 7, the squares modulo 8 are 0, 1, 4, 1, 0, 1, 4, 1, so x² ≡ 1 holds for x = 1, 3, 5 and 7. Thus there are four solutions, which shows that modulo a composite modulus a quadratic congruence can have more than two solutions.
Every perfect square is congruent to 0 or 1 modulo 4, since even squares give 0 and odd squares give 1. A number congruent to 2 modulo 4 therefore cannot be a square. For instance 6, 10 and 14 are all nonsquares.
Since gcd(6, 15) = 3 divides 9, solutions exist and form three classes modulo 15. Dividing the congruence and modulus by 3 gives 2x ≡ 3 (mod 5), and multiplying by the inverse 3 of 2 modulo 5 gives x ≡ 9 ≡ 4 (mod 5).
Wilson's theorem says (p−1)! ≡ −1 (mod p) for every prime p. The proof pairs each residue a with its inverse modulo p; all pairs cancel except 1 and −1, whose product is −1.
The converse of Fermat's little theorem fails in general: a composite n can satisfy 2^(n−1) ≡ 1 (mod n). The number 2047 is such a base-2 pseudoprime, so a Fermat test with base 2 passes it even though 2047 = 23·89 is composite.
Divide 504 repeatedly by primes: 504 = 8 · 63 = 2³ · 3² · 7. Since 2, 3, and 7 are all prime and the exponents are maximal, this is the complete factorization.
The Euclidean algorithm gives 252 = 2·105 + 42, then 105 = 2·42 + 21, then 42 = 2·21 + 0. The last nonzero remainder is 21, so gcd(252, 105) = 21.
We need 5x ≡ 1 (mod 17). Since 5 · 7 = 35 = 2·17 + 1, the product is congruent to 1 modulo 17, so 7 is the inverse of 5.
A reduced residue system modulo 9 contains exactly the classes coprime to 9, of which there are φ(9) = 6: namely 1, 2, 4, 5, 7, and 8. Each is relatively prime to 9, and every other class shares a factor with 9.
The congruence ax ≡ b (mod m) is solvable exactly when gcd(a, m) divides b, and then it has gcd(a, m) solutions. Here gcd(4, 10) = 2 divides 6, so there are 2 solutions, namely x ≡ 4 and x ≡ 9 (mod 10).
Since 36 = 2² · 3², the totient formula gives φ(36) = 36 · (1 − 1/2) · (1 − 1/3) = 36 · 1/2 · 2/3 = 12. Equivalently, φ(4) · φ(9) = 2 · 6 = 12.
By Fermat's little theorem, 3¹⁰ ≡ 1 (mod 11) since 11 is prime and does not divide 3. Because 45 = 4·10 + 5, we get 3⁴⁵ ≡ 3⁵ = 243 ≡ 1 (mod 11).
Candidates congruent to 1 modulo 4 are 1, 5, 9, 13, 17, and so on. Among these, 17 is the smallest that is congruent to 2 modulo 5, since 17 = 3·5 + 2. The Chinese remainder theorem guarantees a unique solution modulo 20.
Since gcd(7, 100) = 1, Euler's theorem with φ(100) = 40 gives 7⁴⁰ ≡ 1 (mod 100). But 7²⁰ ≡ 1 (mod 100) as well, because 7⁴ = 2401 ≡ 1 (mod 100) and 20 is a multiple of 4, so 7²⁰ = (7⁴)⁵ ≡ 1.
The nonzero quadratic residues modulo 7 are 1, 2, and 4, obtained by squaring 1, 2, and 3. Since 2 is among them (3² = 9 ≡ 2), the congruence x² ≡ 2 (mod 7) has the solutions x ≡ 3 and x ≡ 4.
Squaring the residues modulo 7 gives the set {1, 2, 4}, so 3 is a quadratic nonresidue and (3/7) = −1. This agrees with Euler's criterion, since 3³ = 27 ≡ 6 ≡ −1 (mod 7).
Fermat's little theorem gives 3¹² ≡ 1 (mod 13). Since 2024 = 12·168 + 8, we get 3²⁰²⁴ ≡ 3⁸ = 6561, and 6561 = 504·13 + 9, so the remainder is 9.
Checking values in order: φ(15) = 8 since 15 = 3 · 5 and φ(15) = 2 · 4 = 8. All smaller n have φ(n) ≠ 8, so 15 is the least solution.
Since 17 is prime and does not divide 2, Fermat's little theorem gives 2¹⁶ ≡ 1 (mod 17). Because 1000 = 16·62 + 8, we need 2⁸ = 256, and 256 = 15·17 + 1, so the remainder is 1.
Euler's theorem states that when a is coprime to n, raising a to the power φ(n) yields 1 modulo n. The coprimality hypothesis is essential; without it the conclusion can fail, as with a = 2 and n = 4, where 2² = 4 ≡ 0 (mod 4).
The rationals form an abelian group under addition: the sum of two rationals is rational, addition is associative and commutative, 0 is the identity, and every rational q has inverse −q. No further restriction such as common denominators is needed.
Multiplying powers with the same base adds exponents, giving x^8, and dividing subtracts the exponent of the divisor, giving x^(8−2) = x^6. The restriction x ≠ 0 keeps the division valid.
The quadratic factors as (x − 2)(x − 3), since the two numbers multiplying to 6 and summing to −5 are −2 and −3. Hence x − 2 divides the polynomial exactly.
Expanding gives 3x − 12 = 2x + 5. Subtracting 2x from both sides gives x − 12 = 5, and adding 12 gives x = 17. Checking: 3(13) = 39 and 2(17) + 5 = 39.
Slope equals the change in y divided by the change in x: (13 − 5)/(6 − 2) = 8/4 = 2. The line rises two units vertically for every one unit horizontally.
The logarithm asks for the exponent to which the base must be raised: 2^5 = 32, so log₂32 = 5. Logarithms and exponentials are inverse operations.
The expression is a difference of squares a^2 − b^2 with a = 3x and b = 5, which factors as (a − b)(a + b) = (3x − 5)(3x + 5). Expanding confirms 9x^2 − 25.
Subtracting 1 gives −2x < 8. Dividing by −2 reverses the inequality, giving x > −4. Reversing the direction when dividing by a negative number is essential.
Substituting x = −2 gives 2(4) − 3(−2) + 1 = 8 + 6 + 1 = 15. Care is needed because squaring a negative gives a positive and subtracting a negative adds.
A quadratic has exactly one real root when its discriminant b^2 − 4ac equals zero. Here k^2 − 36 = 0, so k^2 = 36 and k = ±6.