Question

Find the largest number which divides 248 and 1032 leaving remainder 8 in each case.


Since the ans is 16
but i don't know how to solve . plzz solve it and give
Quick plzzz

Answer 1

Given that the largest number when divides 248 and 1032, the remainder is 8 in each case.

248 - 8 = 240 and 1032 - 8 = 1024 are completely divisible by the required number.

Therefore, it is the HCF of 240 and 1024.

Prime factorization of 240 = 2 * 2 * 2 * 2 * 3 * 5

Prime factorization of 1024 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2.


HCF(240,1024) = 2 * 2 * 2 * 2

                          = 16.



Therefore the largest number which divides 248 and 1032 leaving remainder 8 in each case = 16.



Hope this helps!

Answer 2

Answer:

The required number is 16

Step-by-step explanation:

Each time when the number divides 248 and 1032 it leaves remainder 8 and 5 respectively.

So, the number exactly divides (248 - 8) and (1032 - 8) ⇒ the number divides 240 and 1024

So, the required number which when divides 248 and 1032 and leaves remainder 8 in each case will be HCF (240, 1024)

Now, to find HCF (240, 1024) : Find the prime factors of 240 and 1024

240 = 2 × 2 × 2 × 2 × 3 × 5

1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

So, we can see the common factor in both the numbers are 2 × 2 × 2 × 2

⇒ HCF = 16

So, the greatest number which when divides 248 and 1032 leaves remainder 8 in each case will be 16