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Prime Factorization - Pre Algebra

  1. Two numbers MM and NN satisfy the equation M⋅N=26⋅34⋅52M \cdot N = 2^6 \cdot 3^4 \cdot 5^2 and GCF(M,N)=22⋅31\text{GCF}(M, N) = 2^2 \cdot 3^1. What is the value of LCM(M,N)\text{LCM}(M, N)?
    Answer: 24⋅33⋅522^4 \cdot 3^3 \cdot 5^2 We use the fundamental relationship: M⋅N=GCF(M,N)⋅LCM(M,N)M \cdot N = \text{GCF}(M, N) \cdot \text{LCM}(M, N). Given M⋅N=26⋅34⋅52M \cdot N = 2^6 \cdot 3^4 \cdot 5^2 and GCF(M,N)=22⋅31\text{GCF}(M, N) = 2^2 \cdot 3^1, we can find: LCM(M,N)=M⋅NGCF(M,N)=26⋅34⋅5222⋅31=26−2⋅34−1⋅52=24⋅33⋅52\text{LCM}(M, N) = \frac{M \cdot N}{\text{GCF}(M, N)} = \frac{2^6 \cdot 3^4 \cdot 5^2}{2^2 \cdot 3^1} = 2^{6-2} \cdot 3^{4-1} \cdot 5^2 = 2^4 \cdot 3^3 \cdot 5^2. Choice A is correct. Choice B incorrectly gives the product M⋅NM \cdot N. Choice C incorrectly adds exponents instead of subtracting. Choice D has incorrect exponents for all primes.
  2. Consider all positive integers of the form 2x⋅3y⋅7z2^x \cdot 3^y \cdot 7^z where xx, yy, and zz are non-negative integers and x+y+z=4x + y + z = 4. How many such integers are there?
    Answer: 1515 We need to count the number of non-negative integer solutions to x+y+z=4x + y + z = 4. This is a classic "stars and bars" combinatorics problem. The number of ways to distribute 4 identical objects into 3 distinct bins is (4+3−13−1)=(62)=15\binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15. Alternatively, we can enumerate: (4,0,0), (3,1,0), (3,0,1), (2,2,0), (2,1,1), (2,0,2), (1,3,0), (1,2,1), (1,1,2), (1,0,3), (0,4,0), (0,3,1), (0,2,2), (0,1,3), (0,0,4). Counting these gives 15 solutions. Choice A (10) incorrectly assumes some constraint on the variables. Choice C (20) might result from incorrectly using (72)\binom{7}{2}. Choice D (25) might result from incorrectly calculating 525^2.
  3. A machine produces widgets in batches. The first batch contains 22⋅312^2 \cdot 3^1 widgets, the second batch contains 21⋅332^1 \cdot 3^3 widgets, and the third batch contains 23⋅522^3 \cdot 5^2 widgets. If all widgets from the three batches are to be packed into identical boxes with no widgets left over, what is the largest number of widgets that can be placed in each box?
    Answer: 22 To pack all widgets into identical boxes with none left over, the number of widgets per box must be a divisor of each batch size. This means we need the GCF of the three batch sizes. Batch 1: 22⋅31=122^2 \cdot 3^1 = 12 widgets. Batch 2: 21⋅33=542^1 \cdot 3^3 = 54 widgets. Batch 3: 23⋅52=2002^3 \cdot 5^2 = 200 widgets. To find GCF(12, 54, 200), we use the minimum exponent for each prime: For prime 2: min⁡(2,1,3)=1\min(2, 1, 3) = 1. For prime 3: min⁡(1,3,0)=0\min(1, 3, 0) = 0 (since 200 has no factor of 3). For prime 5: min⁡(0,0,2)=0\min(0, 0, 2) = 0 (since 12 and 54 have no factor of 5). Therefore, GCF = 21⋅30⋅50=22^1 \cdot 3^0 \cdot 5^0 = 2. Choice B (4) would not divide 54 evenly since 54 = 2¹ × 3³, and 4 = 2². Choice C (6) would not divide 200 since 200 has no factor of 3. Choice D (12) would not divide 200 evenly since 200 has no factor of 3.

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