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# If the least prime factor for number ‘a’ is 3 and the least prime factor for number ‘b’ is 7. Then what will least prime factor of number (a+b) is:(a) 2(b) 3(C) 5(d) 11

Last updated date: 16th May 2024
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Hint: we use the fact that 2 is an even prime number and can divide all even numbers. So, it makes us remain with odd numbers as only possibility for numbers ‘a’ and ‘b’ because ‘3’ and ‘7’ are odd numbers. Now using these properties for numbers ‘a’ and ‘b’, we take assumption in general form for numbers ‘a’ and ‘b’. Once we find general form for ‘a’ and ‘b’ we find the least prime factor for (a+b).

According to the problem, we have a number ‘a’ whose least prime factor is 3 and number ‘b’ whose least prime factor is 7. We need to find the least prime factor of number (a+b).
We first find the characteristics for numbers ‘a’ and ‘b’ and then solve for least prime factor for number (a+b).
Now, if we check on number ‘a’, the least prime factor is 3. Here 3 is an odd number and it can divide both even and odd numbers. But ‘a’ cannot be an even number as all even numbers are divisible by ‘2’ which is prime number and less than ‘3’. So, ‘a’ is an odd number.
We know an odd number can be represented as $2p-1(p\ge 1)$.
We assume $c=2n-1$ to represent an odd number. But we need odd numbers that are divisible by 3 and every number of the form $2n-1$ is not divisible by 3. So, we multiply $2n-1$ with 3.
We have $a=3c$.
$\Rightarrow a=3\left( 2n-1 \right)$.
$\Rightarrow a=6n-3$ ---(1).
So, $a=6n-3$ is the general form of the number that has least prime factor ‘3’.
Now, if we check on number ‘b’, the least prime factor is 7. Here 7 is an odd number and it can divide both even and odd numbers. But ‘b’ cannot be an even number as all even numbers are divisible by ‘2’ which is prime number and less than ‘7’. So, ‘b’ is an odd number.
We assume $d=2m-1$ to represent an odd number. But we need odd numbers that are divisible by 3 and every number of the form $2m-1$ is not divisible by 7. So, we multiply $2m-1$ with 7.
We have $b=7d$.
$\Rightarrow b=7\left( 2m-1 \right)$.
$\Rightarrow b=14m-7$---(2).
Now we find the least prime factor for number (a+b). Using equations (1) and (2) we get,
$\Rightarrow a+b=\left( 6n-3 \right)+\left( 14m-7 \right)$.
$\Rightarrow a+b=6n+14m-10$.
$\Rightarrow a+b=2\times \left( 3n+7m-5 \right)$---(3).
From equation (3), we can see that number (a+b) is clearly divisible by 2 which is prime.
So, the least prime factor for number (a+b) is 2.

So, the correct answer is “Option A”.

Note: Here we should not add 3 and 7 directly to find the least prime factor for (a+b), this may provide us with answer 2 but it doesn’t happen every time with other numbers. Always remember that odd prime numbers can divide both even and odd numbers but even numbers can divide only even numbers. 2 is an even prime number and can divide all even numbers to proceed through the problem.