What is the greatest common divisor of 84 and 126?
- 14
- 21
- 42 Correct answer.
- 63
Explanation
Prime factorizations: 84 = 2^2·3·7 and 126 = 2·3^2·7. The GCD takes the smallest power of each shared prime: 2·3·7 = 42.
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Prime factorizations: 84 = 2^2·3·7 and 126 = 2·3^2·7. The GCD takes the smallest power of each shared prime: 2·3·7 = 42.
12 = 2^2·3 and 18 = 2·3^2, so the LCM takes the largest power of each prime: 2^2·3^2 = 36. Equivalently, LCM = (12·18)/GCD(12,18) = 216/6 = 36.
A number is prime if it has no divisor other than 1 and itself; it suffices to check primes up to its square root. Since √97 < 10, testing 2, 3, 5 and 7 shows none divides 97, so 97 is prime.
Since 2^3 = 8 ≡ 1 (mod 7), the powers of 2 repeat with period 3. Writing 10 = 3·3 + 1 gives 2^10 ≡ (2^3)^3·2 ≡ 1·2 ≡ 2 (mod 7).
Factor 360 = 2^3·3^2·5. The number of positive divisors is the product of one more than each exponent: (3+1)(2+1)(1+1) = 24.
The units digits of powers of 3 cycle 3, 9, 7, 1 with period 4. Since 100 is divisible by 4, 3^100 has the same units digit as 3^4, namely 1.
Numbers congruent to 3 mod 5 are 3, 8, 13, 18, ... Checking these mod 7 gives 3, 1, 6, 4, so 18 is the first to satisfy both conditions.
Compute 4^10 = (2^10)^2 = 1024^2 = 1,048,576. Adding the digits gives 1+0+4+8+5+7+6 = 31, so rechecking: 4^5 = 1024, 4^10 = 1,048,576, whose digits sum to 31.
A number is divisible by 9 exactly when its digit sum is divisible by 9. The digits of 4,536 sum to 4+5+3+6 = 18, which is divisible by 9.
Dividing repeatedly by the smallest primes: 1260 = 2·630 = 2^2·315 = 2^2·3·105 = 2^2·3^2·35 = 2^2·3^2·5·7.
Since 5 ≡ -1 (mod 6), we get 5^2023 ≡ (-1)^2023 ≡ -1 ≡ 5 (mod 6). The odd exponent keeps the result at 5 rather than 1.
1,000 = 2^3·5^3, so n! must contain three factors of 5. The count of factors of 5 in n! is ⌊n/5⌋+⌊n/25⌋+..., reaching 3 first at n = 15, where 2s are abundant.
Modulo 5, 3^4 ≡ 1 and 4^4 ≡ 1 by Fermat's little theorem, and 100 is a multiple of 4, so 3^100 ≡ 1 and 4^100 ≡ 1. The sum is therefore ≡ 2 (mod 5).
Euler's totient for 100 = 2^2·5^2 is φ(100) = 100·(1-1/2)·(1-1/5) = 100·(1/2)·(4/5) = 40. So 40 numbers in the range share no factor with 100.
The sum of the first n odd integers is n^2, so for n = 50 it equals 2,500. This follows because each odd number 2k-1 sums to n^2.
Powers of 7 have units digits cycling 7, 9, 3, 1 with period 4. Since 2024 = 4·506, the exponent is a multiple of 4, so the units digit is 1.
Modular multiplication allows us to multiply residues: 3·5 = 15 ≡ 7 (mod 8), since 15 - 8 = 7.
By Fermat's little theorem, 10^12 ≡ 1 (mod 13). Since 50 = 12·4 + 2, we get 10^50 ≡ 10^2 = 100 ≡ 9, so recomputing: 100 mod 13 = 9, hence the remainder is 9, which is not listed; the closest valid residue is 12 only if the exponent were 6 mod 12.
If n = p^a, the divisor count is a+1; three divisors requires a = 2, so n = p^2 for a prime p. A number with two distinct prime factors would have at least four divisors.
If n ≡ 5 (mod 6), then n + 10 ≡ 5 + 10 = 15 ≡ 3 (mod 6), since 15 = 2·6 + 3. Adding 10 is the same as adding 4 modulo 6, and 5 + 4 = 9 leaves remainder 3.
Distribute the 3 over the parentheses to get 6x - 15, then add 4x. Combining like terms gives 6x + 4x - 15 = 10x - 15.
This is a difference of two squares, a² - b² = (a - b)(a + b), with a = x and b = 3. Applying the identity gives (x - 3)(x + 3).
First square a: (-2)² = 4, then multiply by 2 to get 8. Subtracting 3b = 12 gives 8 - 12 = -4... careful: 8 - 12 = -4, so the value is -4.
When a power is raised to a power, the exponents are multiplied: (xᵐ)ⁿ = x^(m·n). Here 3 · 4 = 12, so the result is x¹².
Divide the coefficients: 6/3 = 2. For each variable, subtract the denominator exponent from the numerator exponent: x^(5-2) = x³ and y^(2-1) = y.
Subtract 3x from both sides to get 2x - 7 = 9, then add 7 to get 2x = 16. Dividing by 2 gives x = 8.
Multiply each term of the first bracket by each term of the second: x·x + x·(-2) + 4·x + 4·(-2). This gives x² - 2x + 4x - 8, and combining -2x + 4x yields x² + 2x - 8.
The numerator is a difference of squares: x² - 4 = (x - 2)(x + 2). Cancelling the common factor (x - 2) leaves x + 2.
Substitute x = 3 into the equation: 2(3) + 3y = 12, so 6 + 3y = 12. Subtracting 6 gives 3y = 6, hence y = 2.
We need two numbers whose product is 6 and whose sum is -5. The numbers -2 and -3 satisfy this, so the factorization is (x - 2)(x - 3).
Factor the quadratic as (x - 3)(x - 4) = 0, since -3 and -4 multiply to 12 and add to -7. Setting each factor to zero gives x = 3 or x = 4.
Raise each factor to the third power: 2³ = 8, (x²)³ = x⁶, and (y³)³ = y⁹. The result is 8x⁶y⁹.
Multiply both sides by 2 to get 2y = 3x - 5, then add 5 to get 2y + 5 = 3x. Dividing by 3 gives x = (2y + 5)/3.
The fractions share the denominator x + 1, so add the numerators: x + 1 over x + 1. Since x + 1 divided by x + 1 equals 1 for x ≠ -1, the expression simplifies to 1.
Let the numbers be a and b with a + b = 20 and a - b = 6. Adding the equations gives 2a = 26, so a = 13.
Adding the two equations eliminates y: 3x = 12, so x = 4. Substituting into x - y = 1 gives 4 - y = 1, so y = 3.
The discriminant is b² - 4ac with a = 2, b = -4, c = 5. This gives (-4)² - 4(2)(5) = 16 - 40 = -24.
Factor both parts: the numerator is (x - 3)(x + 2) and the denominator is (x - 3)(x + 3). Cancelling the common factor (x - 3) leaves (x + 2)/(x + 3).
Square both sides: (x + 1/x)² = x² + 2 + 1/x² = 9. Subtracting 2 gives x² + 1/x² = 7.
Subtract 3 from both sides to get -2x < 4. Dividing by -2 reverses the inequality sign, giving x > -2.
By the remainder theorem, the remainder is P(3) = 27 - 36 + 3 + 6 = 0. So x - 3 is also a factor of the polynomial.
A quadratic has exactly one real root when its discriminant b² - 4ac is zero. Here k² - 4(1)(9) = 0, so k² = 36 and k = ±6.
Use the identity a³ + b³ = (a + b)³ - 3ab(a + b). Substituting gives 125 - 3(6)(5) = 125 - 90 = 35.
For any x between -2 and 2, the distances to 2 and to -2 add to exactly 4, since the points lie between the two centres. Hence every x in [-2, 2] satisfies the equation, giving infinitely many solutions.
Distribute the 4 and the -2: 4(2x - 3) = 8x - 12 and -2(x - 5) = -2x + 10. Then combine like terms: (8x - 2x) + (-12 + 10) = 6x - 2.
First factor out the common factor 3: 3(x² - 4). Then recognize x² - 4 as a difference of squares, x² - 2² = (x - 2)(x + 2). So the complete factorization is 3(x - 2)(x + 2).
Cross-multiply to get 2(x + 2) = 3(x - 1). Expand: 2x + 4 = 3x - 3. Rearrange: 4 + 3 = 3x - 2x, so x = 7. Check that x = 7 does not make any denominator zero, which it does not.
Write both sides with base 2: 8 = 2³, so 8^(x-1) = (2³)^(x-1) = 2^(3x-3). Equate exponents: x + 1 = 3x - 3. Solve: 4 = 2x, so x = 2.
A perfect square trinomial has the form (x + k)² = x² + 2kx + k². Here k² = 16, so k = 4 or k = -4. Then b = 2k, giving b = 8 or b = -8.
A quadratic has two distinct real roots when its discriminant is positive. Here the discriminant is k² - 4(k + 3) = k² - 4k - 12. Solve k² - 4k - 12 > 0: the roots of the corresponding equation are k = -2 and k = 6, and the parabola opens upward, so the expression is positive when k < -2 or k > 6.