CAT — Factors and Multiples

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

Prime factorize the given numbers.
168 = 2^3 × 3 × 7
1134 = 2 × 3^4 × 7
Therefore,
1134^n = 2^n × 3^(4n) × 7^n
For 168 to be a factor of 1134^n, the powers of each prime in 1134^n must be at least those in 168.
Compare the exponents:
For 2: n ≥ 3
For 3: 4n ≥ 1 → n ≥ 1/4
For 7: n ≥ 1
The largest requirement is n = 3
Now, 1134^3 = 2^3 × 3^12 × 7^3
Next, 168^m = 2^(3m) × 3^m × 7^m
For 1134^3 to be a factor of 168^m, compare the exponents.
For 2: 3m ≥ 3 → m ≥ 1
For 3: m ≥ 12
For 7: m ≥ 3
The largest requirement is m = 12
Therefore, m + n = 12 + 3 = 15
Answer:
B. 15

Non-zero digits excluding perfect squares (1, 4, 9): allowed digits are 2, 3, 5, 6, 7, 8. Among these, prime digits are 2, 3, 5, 7 and composite digits are 6, 8. Since only one digit is prime, the number must contain both composites (6 and 8) and exactly one prime.

To minimise N: hundreds digit = 6, tens digit = 2 (smallest prime), units digit = 8. So N = 628.

Prime factorisation: 628 = 2² × 157 (157 is prime). Number of factors = (2+1)(1+1) = 6.

Step 1: Use the property of divisibility
If k divides two numbers, it also divides any integer linear combination of those numbers.
Given: k | (m + 2n) and k | (3m + 4n)

Step 2: Eliminate n
Multiply the first expression by 2: 2(m + 2n) = 2m + 4n
Subtract from the second: (3m + 4n) − (2m + 4n) = m
Hence: k | m

Step 3: Find another quantity divisible by k
Since k | (m + 2n) and k | m:
Subtract: (m + 2n) − m = 2n
Hence: k | 2n

Step 4: Conclusion
k is a common divisor of m and 2n.

Given:
A function f satisfies
f(xy) = f(x)f(y) + f(x) + f(y)
for all natural numbers x and y.
Also,
f(p) = 1 for every prime number p.
Find the value of
f(160000).

Step 1: Define a new function
Let
g(x) = f(x) + 1
Then,
g(xy)
= f(xy) + 1
= f(x)f(y) + f(x) + f(y) + 1
= (f(x) + 1)(f(y) + 1)
= g(x)g(y)
Thus,
g(xy) = g(x)g(y)

Step 2: Find g for prime numbers
Since
f(p) = 1,
we have
g(p) = 2
for every prime p.

Step 3: Prime factorize 160000
160000
= 160 × 1000
= (2⁵ × 5) × (2³ × 5³)
= 2⁸ × 5⁴
Since g is multiplicative,
g(160000)
= g(2)⁸ × g(5)⁴
= 2⁸ × 2⁴
= 2¹²
= 4096

Step 4: Find f(160000)
f(160000)
= g(160000) − 1
= 4096 − 1
= 4095

Answer:
A. 4095

Step 1: Understand the condition A number has exactly two distinct factors other than 1 and itself if it has exactly 4 divisors in total. Numbers with exactly 4 divisors are of two types: p³ where p is prime, or p × q where p and q are distinct primes. Step 2: Find numbers of the form p³ less than 50 2³ = 8, 3³ = 27 → 2 such numbers. Step 3: Find numbers of the form p × q less than 50 Using prime 2: 6, 10, 14, 22, 26, 34, 38, 46 Using prime 3: 15, 21, 33, 39 Using prime 5: 35 Using prime 7: 7 × 11 = 77 > 50, stop. Total: 13 such numbers. Step 4: Total count = 2 + 13 = 15

Step 1: Write the general form of a divisor. The given number is 2⁶ × 3⁵ × 5³ × 7². A divisor is of the form 2ᵃ × 3ᵇ × 5ᶜ × 7ᵈ where 0 ≤ a ≤ 6, 0 ≤ b ≤ 5, 0 ≤ c ≤ 3, 0 ≤ d ≤ 2.

Step 2: Use the condition that the divisor is of the form 3r + 1. A number of the form 3r + 1 leaves remainder 1 when divided by 3. If b ≥ 1, then the divisor is divisible by 3. Hence, b = 0. Now the divisor becomes 2ᵃ × 5ᶜ × 7ᵈ.

Step 3: Find the remainder modulo 3. Modulo 3: 2 ≡ −1, 5 ≡ −1, 7 ≡ 1. Therefore, 2ᵃ × 5ᶜ × 7ᵈ ≡ (−1)ᵃ × (−1)ᶜ × 1ᵈ = (−1)^(a+c). For the remainder to be 1, a + c must be even.

Step 4: Count the valid values of a and c. Possible values of a: 0 to 6. Even values: 0, 2, 4, 6 → 4 choices. Odd values: 1, 3, 5 → 3 choices. Possible values of c: 0 to 3. Even values: 0, 2 → 2 choices. Odd values: 1, 3 → 2 choices. For a + c to be even: both even = 4 × 2 = 8 pairs; both odd = 3 × 2 = 6 pairs. Total valid pairs = 8 + 6 = 14.

Step 5: Choose the value of d. Since 7 ≡ 1 (mod 3), the value of d does not affect the remainder. Possible values of d are 0, 1 and 2, giving 3 choices. Total number of divisors = 14 × 3 = 42.

Given:
m and n are natural numbers with
n > 1
and
mⁿ = 2²⁵ × 3⁴⁰.
Find the value of
m − n.

Step 1: Find the possible value of n
Since m is an integer, the exponents of every prime in mⁿ must be divisible by n.
Therefore,
n must divide both 25 and 40.
The common divisors of 25 and 40 are
1 and 5.
Since
n > 1,
we get
n = 5.

Step 2: Find m
m
= (2²⁵ × 3⁴⁰)^(1/5)
= 2⁵ × 3⁸
Now,
2⁵ = 32
3⁸ = 6561
Therefore,
m = 32 × 6561
= 209952

Step 3: Find m − n
m − n
= 209952 − 5
= 209947

Answer:
D. 209947