Question

find the greatest number which can divide 257 and 329 so that to leave a remainder 5 in each case

Answer 1

Answer:

The required number is 36.

Step-by-step explanation:

Each time when the number divides 257 and 329 it leaves remainder 5

So, the number exactly divides (257 - 5) and (329 - 5) ⇒ the number divides 252 and 324

So, the required number which when divides 257 and 329 and leaves remainder 5 will be HCF (252,324)

Now, to find HCF (252,324) : Find the prime factors of 252 and 325

252 = 2 × 2 × 3 × 3 × 7

325 = 2 × 2 × 3 × 3 × 3 × 3

So, we can see the common factor in both the numbers are 2, 2, 3, 3

⇒ 2 × 2 × 3 × 3 = 36

So, the greatest number which when divides 257 and 329 leaves remainder 5 each time is 36

 

Answer 2

Answer:

Let x=257

y=329

Given R=5

⇒x−5=257−5⇒252

⇒y−5=329−5⇒324

Factorize 252 & 324

252=2

2

×3

2

×7

324=2

2

×3

4

G.C.F⇒2

2

×3

2

=4×9=36.