CAT — HCF and LCM
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Let a, b, m and n be natural numbers such that a > 1 and b > 1. If aᵐbⁿ = 144¹⁴⁵, then the largest possible value of n − m is
580
290
289
579
579
Step 1: Prime factorize 144
144 = 2⁴ × 3²
Therefore, 144¹⁴⁵ = (2⁴ × 3²)¹⁴⁵ = 2⁵⁸⁰ × 3²⁹⁰
So, aᵐ × bⁿ = 2⁵⁸⁰ × 3²⁹⁰
Step 2: Observe how to maximize (n − m)
To maximize (n − m): make m as small as possible and n as large as possible.
The smallest possible value of m is 1.
Step 3: Reserve the smallest possible factor for a
Take a = 2, m = 1. Then aᵐ = 2.
Remaining factor = 2⁵⁷⁹ × 3²⁹⁰
gcd(579, 290) = 1, so this choice does not maximize n.
Step 4: Try reserving one factor of 12
Take a = 12, m = 1. Then aᵐ = 2² × 3.
Remaining factor = 2⁵⁷⁸ × 3²⁸⁹
gcd(578, 289) = 289, so n = 289.
This gives n − m = 289 − 1 = 288. Still not maximum.
Step 5: Best possible construction
The maximum value is obtained by choosing the factorization such that m = 1 and n = 580.
Therefore, n − m = 580 − 1 = 579
Let aₙ and bₙ be two sequences such that aₙ = 13 + 6(n − 1) and bₙ = 15 + 7(n − 1) for all natural numbers n. Then, the largest three digit integer that is common to both these sequences, is
aₙ = 6n + 7 → sequence: 13, 19, 25, ... (numbers ≡ 13 mod 6 ≡ 1 mod 6) bₙ = 7n + 8 → sequence: 15, 22, 29, ... (numbers ≡ 15 mod 7 ≡ 1 mod 7) Common terms satisfy: x ≡ 1 (mod 6) and x ≡ 1 (mod 7) → x ≡ 1 (mod 42) Every common term is of the form x = 42k + 1. Largest three-digit value: 42k + 1 ≤ 999 → 42k ≤ 998 → k ≤ 23 x = 42 × 23 + 1 = 966 + 1 = 967
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