Let n be the least positive integer such that 180 is a factor of 270^n. If m is the least positive integer such that 270^n is a factor of 180^m, find m+n.
Answer and solution
Correct answer: B
Solution
Answer: B — 5
1. Prime factorizations: 180=2²×3²×5 and 270=2×3³×5.
2. For 180 to divide 270ⁿ, require n≥2,3n≥2, n≥1. Therefore the least n is 2.
3. Now 270²=2²×3⁶×5² must divide 180ᵐ=2²ᵐ×3²ᵐ×5ᵐ.
4. Need 2m≥2,2m≥6, m≥2. Least m=3. Hence m+n=5, option B.
Why this method works:
For divisibility, each prime exponent in the divisor must be no larger than the corresponding exponent in the dividend. The strongest of these inequalities determines the smallest exponent; checking only one arbitrary prime can miss the constraint.
Independent check and common trap:
For n=1 the power of 2 is too small, while n=2 works. Then m=2 gives only 3⁴, too small for the required 3⁶; m=3 supplies 2⁶,3⁶,5³ and works for every prime. Thus both minima are independently established.
Apply: 180=2²3²5 and 270=2·3³5. First the scarce 2 forces n=2; then the required 3⁶ forces m=3. Sum=5.
Check: For n=1 the power of 2 is too small, while n=2 works. Then m=2 gives only 3⁴, too small for the required 3⁶; m=3 supplies 2⁶,3⁶,5³ and works for every prime. Thus both minima are independently established.
1. Prime factorizations: 180=2²×3²×5 and 270=2×3³×5.
2. For 180 to divide 270ⁿ, require n≥2,3n≥2, n≥1. Therefore the least n is 2.
3. Now 270²=2²×3⁶×5² must divide 180ᵐ=2²ᵐ×3²ᵐ×5ᵐ.
4. Need 2m≥2,2m≥6, m≥2. Least m=3. Hence m+n=5, option B.
Why this method works:
For divisibility, each prime exponent in the divisor must be no larger than the corresponding exponent in the dividend. The strongest of these inequalities determines the smallest exponent; checking only one arbitrary prime can miss the constraint.
Independent check and common trap:
For n=1 the power of 2 is too small, while n=2 works. Then m=2 gives only 3⁴, too small for the required 3⁶; m=3 supplies 2⁶,3⁶,5³ and works for every prime. Thus both minima are independently established.
G Strategy shortcut
Spot: HCF and LCM / Prime Factorisation.Apply: 180=2²3²5 and 270=2·3³5. First the scarce 2 forces n=2; then the required 3⁶ forces m=3. Sum=5.
Check: For n=1 the power of 2 is too small, while n=2 works. Then m=2 gives only 3⁴, too small for the required 3⁶; m=3 supplies 2⁶,3⁶,5³ and works for every prime. Thus both minima are independently established.
G Strategy: Spot: HCF and LCM / Prime Factorisation. Apply: 180=2²3²5 and 270=2·3³5. First the scarce 2 forces n=2; then the required 3⁶ forces m=3. Sum=5. Check: For n=1 the power of 2 is too small, while n=2 works. Then m=2 gives only 3⁴, too small for the required 3⁶; m=3 supplies 2⁶,3⁶,5³ and works for every prime. Thus both minima are independently established.