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+Question:
+The greatest common divisor of positive integers m and n is 6. The least common multiple of m and n is 126. What is the least possible value of m + n? Answer: 60
You are not allowed to use any code. The problem is: {Question}.
+We know that for any two positive integers m and n, the product of their greatest common divisor (GCD) and least common multiple (LCM) is equal to the product of the two numbers. This can be written as: GCD(m, n) ⋅ LCM(m, n) = m ⋅ n
+Given that GCD(m, n) = 6 and LCM(m, n) = 126, we can substitute these values into the equation:
+6 ⋅ 126 = m ⋅ n
+Solving this gives us m ⋅ n = 756.
+Now, we need to find the pair of factors of 756 that add up to the smallest sum, as we want to find the least possible value of m + n.
+The pairs of factors of 756 are (1,756), (2,378), (3,252), (4,189), (6,126), (7,108), (9,84), (12,63), (14,54), (18,42), (21,36), (27,28).
+We can see that the pair (21,36) gives the smallest sum, which is 21 + 36 = 57.
+So, the least possible value of m + n is 57.
+Therefore, the solution to the
+problem is $\boxed{57}$.
+You can use code only 1 time. The problem is: {Question}.
+First, we should recall that for any two positive integers m and n, their greatest common divisor (gcd) and least common multiple (lcm) satisfy the relationship: m ⋅ n = gcd(m, n) ⋅ lcm(m, n)
+In this problem, we’re given that gcd(m, n) = 6 and lcm(m, n) = 126. We can substitute these values into the formula to get: m ⋅ n = 6 ⋅ 126
+We can solve this equation for one variable and then use the fact that m and n are integers to generate possible pairs (m, n).
+After we have the pairs (m, n), we can find the pair that minimizes m + n.
+import sympy as sp
+gcd,lcm = 6,126
+product = gcd * lcm
+pairs = [(d, product // d) for d in sp.divisors(product) if d < = product // d]
+min_pair = min(pairs, key=sum)
+min_pair, sum(min_pair)
+>>> ((27, 28), 55)
The pair (m, n) that minimizes m + n is (27, 28), and the minimum possible value of m + n is $\boxed{55}$.
+Solve the problem and put your answer in \\boxed{}. The problem is: {Question}.
One important property relating greatest common divisor (GCD) and least common multiple (LCM) is that GCD(m, n) × LCM(m, n) = m × n Given in the problem, we can get 6 × 126 = m × n Let’s compute the right hand side of the equation first.
+gcd,lcm = 6, 126
+gcd * lcm
+>>> 756
The product of m and n is 756.
+Since the GCD of m and n is 6, both m and n must be multiples of 6.
+Let’s find all the pairs of multiples of 6 that multiply to 756. We’ll do this by finding all the factors of 756 that are multiples of 6.
+
+>>> [(6, 126), (12, 63), (18, 42)]
The pairs of factors of 756 that are multiples of 6 are:
+1. (6, 126) 2. (12, 63) 3. (18, 42)
+We want to find the pair with the least sum. Let’s compute the sums for each pair,identify the smallest one.
+sums = [(pair, sum(pair)) for pair in factor_pairs]
+min(sums, key=lambda x: x[1])
+>>> ((18, 42), 60)
The pair with the smallest sum is (18, 42), and the sum is 60.So, the least possible value is $\boxed{60}$.
+2
+Question:
+What is the smallest positive integer solution to $(\frac{100,000,000}{x^4}) < 1$?
Answer: 101
+Breaking this down into subproblems, the first step is to rearrange the inequality to isolate x.
+We can do this by multiplying both sides of the inequality by x4 and then taking the fourth root of both sides.
+This gives us $x>\sqrt[4]{100000000}$.
Let’s solve for x.
+import math
+root = math.pow(100000000, 1/4)
>>> 100
Since 100 is already a positive integer, the smallest positive integer solution to the inequality $\frac{100,000,000}{x^4} < 1$ is x = 100.
+The final answer is .
We can firstly change the inequality into $x>\sqrt[4]{100000000}$. Then let’s evaluate the fourth root of 100,000,000.
+import math
+root = math.pow(100000000, 1/4)
+smallest_integer = math.ceil(root)
>>> (100.0, 100)
Thus, the smallest positive integer x is 100.
+Let’s verify this solution.
verification = 1e8 / (smallest_integer ** 4) < 1
+>>> False
The smallest integer x should be the next integer greater than 100, not 100 itself.
+Let’s calculate that next integer.
smallest_integer = 100 + 1
+verification = 1e8 / (smallest_integer ** 4) < 1
+
>>> (101, True)
Upon substituting x = 101 back, it holds true, which verifies our solution.
+The final answer is .
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