Question

Find the largest number which divides $ 630 $ and $ 940 $ leaving the remainder $ 6 $ and $ 4 $ .

Answer 1

Hint: In this question we have to find the largest number which divides both the numbers. First we will subtract the remainder from their respective numbers and then we get two different numbers. After that we will find the highest common factor or H.C.F of both the numbers by the method of prime factorisation.

Complete step by step solution:
We have two numbers $ 630 $ and $ 940 $
And the remainders we have to subtract are $ 6 $ and $ 4 $ .
Since dividing $ 630 $ by the required number, the remainder is $ 6 $ . So after subtracting we have,
 $ 630 - 6 = 624 $
Therefore $ 624 $ will be exactly divisible by the required number.
Similarly we will subtract $ 4 $ from $ 940 $ . We have:
 $ 940 - 4 = 936 $
So $ 936 $ will also be exactly divisible by the required number.
Now we have to find the HCF of $ 624 $ and $ 936 $ i.e.
 $ H.C.F(624,936) $
We will now use the prime factorisation of both the numbers. We know that the numbers that are divisible by $ 1 $ and itself are called prime factors.
So we can write $ 624 $ as the factors of
 $ 624 = 2 \times 2 \times 2 \times 2 \times 3 \times 13 $
These are the prime factors of the given number.
Now we will find the prime factors of $ 936 $ i.e.
 $ 936 = 2 \times 2 \times 2 \times 3 \times 3 \times 13 $
So these are the prime factors of the given number.
Now we know that in HCF we take all the common factors within the numbers, it gives us
 $ HCF(624,936) = 2 \times 2 \times 2 \times 3 \times 13 $
On multiplying the numbers it gives,
 $ HCF(624,936) = 312 $ .
Hence the required number is $ 312 $ which divides $ 630 $ and $ 940 $ leaving remainder $ 6 $ and $ 4 $
So, the correct answer is “312”.

Note: As we know that $ 2 $ is the smallest even prime number. We should note that $ 1 $ is not a prime number. For any number to be prime it has two positive factors, one and itself. But in the case of $ 1 $ , it has only one factor i.e. itself. So it is not a prime number. We should note that in prime factorisation, we will always have prime numbers as factors. There is also another property of prime numbers which is that they must be greater than $ 1 $ . We should note that there are two other methods to find factors under this, they are longer methods- Division method and Factor tree method.