Question

The product of two numbers is 3388 and their HCF is 11. Find all the possible pairs of numbers.

Answer 1

Step-by-step explanation:

Let the numbers be a and b

Let a=p*16, b=q*16; where p&q are co-prime positive integers

Now, ab = pq*(16^2) => pq = 28 = 2^2*7

Number of ordered pairs (p,q) such that p and q are co-prime is the same as the number of factors f of pq such that gcd(f,pq/f)=1. This is all factors such that they are either divisible by the largest power of a specific prime factor in prime factorization of pq, or not divisible by that prime factor at all.

Hence, the number of ordered pairs (p,q) is 2^omega(pq) = 2^2 = 4.

omega function here refers to the number of prime factors of a number n:

If p,q is to be considered a set, which the question implies it to be, then we need to add the condition that q > p, and the number of combinations is 4/2 = 2

If we can consider negative numbers as well, then the answer is 2*2 = 4.

As to the exact set of numbers,

Assuming a, b are positive numbers, following combinations exist:

p=1,q=28 => a= 16, b=448

p=4,q=7 => a= 64, b= 112

Rest are just the same numbers in opposite order.

Assuming a,b can be negative numbers, following four combinations exist:

a=16, b= 448; a=-16, b=-448

a=64, b= 224; a=-64, b=-112

Rest are just the same numbers in opposite order.

Answer 2

Answer:

Hey mate

Step-by-step explanation:

Here is ur answer attached.....

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