If 6^(4x) × 9^(3x+6) × 25^y = 27^(2z) × 10^(4x) × 125^(x-2), where x, y and z are natural numbers, find the minimum possible value of x+y+z.
Answer and solution
Correct answer: 36
Solution
Answer: 36
1. Rewrite every base using primes 2,3,5. Left side is 2⁴ˣ×3¹⁰ˣ⁺¹²×5²ʸ. Right side is 2⁴ˣ×3⁶ᶻ×5⁷ˣ⁻⁶.
2. Equate exponents: z=(5x+6)/3 and y=(7x−6)/2.
3. For z to be an integer, x is divisible by 3. For y to be an integer, x is even. Thus x=6k.
4. Then y=21k−3 and z=10k+2. Nonnegative x=0 would give y=−3, so every natural-number convention requires k≥1.
5. x+y+z=37k−1 increases with k. At k=1: (x, y, z)=(6,18,12), total=36. The word minimum makes the answer unique.
Why this method works:
Unique prime factorization permits comparison of the exponents of 2,3 and 5 separately. The powers of 2 already match, leaving two linear relations. Integrality then supplies the crucial divisibility conditions that ordinary real-variable solving would miss.
Independent check and common trap:
At the smallest feasible x=6, y=(42−6)/2=18 and z=(30+6)/3=12. Check prime exponents:2 has 24 on both sides,3 has 72 and 5 has 36. Larger x=6k increases the sum by 37 each time, proving minimality.
Apply: Ignore matching powers of 2. Powers of 3 force x to be a multiple of 3; powers of 5 force x even. Smallest feasible x=6, yielding y=18, z=12.
Check: At the smallest feasible x=6, y=(42−6)/2=18 and z=(30+6)/3=12. Check prime exponents:2 has 24 on both sides,3 has 72 and 5 has 36. Larger x=6k increases the sum by 37 each time, proving minimality.
1. Rewrite every base using primes 2,3,5. Left side is 2⁴ˣ×3¹⁰ˣ⁺¹²×5²ʸ. Right side is 2⁴ˣ×3⁶ᶻ×5⁷ˣ⁻⁶.
2. Equate exponents: z=(5x+6)/3 and y=(7x−6)/2.
3. For z to be an integer, x is divisible by 3. For y to be an integer, x is even. Thus x=6k.
4. Then y=21k−3 and z=10k+2. Nonnegative x=0 would give y=−3, so every natural-number convention requires k≥1.
5. x+y+z=37k−1 increases with k. At k=1: (x, y, z)=(6,18,12), total=36. The word minimum makes the answer unique.
Why this method works:
Unique prime factorization permits comparison of the exponents of 2,3 and 5 separately. The powers of 2 already match, leaving two linear relations. Integrality then supplies the crucial divisibility conditions that ordinary real-variable solving would miss.
Independent check and common trap:
At the smallest feasible x=6, y=(42−6)/2=18 and z=(30+6)/3=12. Check prime exponents:2 has 24 on both sides,3 has 72 and 5 has 36. Larger x=6k increases the sum by 37 each time, proving minimality.
G Strategy shortcut
Spot: Indices and Prime Factorisation.Apply: Ignore matching powers of 2. Powers of 3 force x to be a multiple of 3; powers of 5 force x even. Smallest feasible x=6, yielding y=18, z=12.
Check: At the smallest feasible x=6, y=(42−6)/2=18 and z=(30+6)/3=12. Check prime exponents:2 has 24 on both sides,3 has 72 and 5 has 36. Larger x=6k increases the sum by 37 each time, proving minimality.
G Strategy: Spot: Indices and Prime Factorisation. Apply: Ignore matching powers of 2. Powers of 3 force x to be a multiple of 3; powers of 5 force x even. Smallest feasible x=6, yielding y=18, z=12. Check: At the smallest feasible x=6, y=(42−6)/2=18 and z=(30+6)/3=12. Check prime exponents:2 has 24 on both sides,3 has 72 and 5 has 36. Larger x=6k increases the sum by 37 each time, proving minimality.