Question

4. Find the greatest number which can divide 257 and 329 so as to leave a remainder
5 in each case.

please tell the full method of solving.
​

Answer 1

Answer: 36

Step-by-step explanation:

Since remainder will be 5 we will subtract 5 from each of the numbers:

257 - 5 = 252

329 - 5 = 324

Since we have to find the greatest number which can divide them both, we will have to find the HCF of 252 and 324:

252/2 = 126

126/2 = 63

63/7 = 9

9/3 = 3

3/3 = 1

∴ Prime factors of 252 are: 2 x 2 x 7 x 3 x 3

324/2 = 162

162/2 = 81

81/3 = 27

27/3 = 9

9/3 = 3

3/3 = 1

∴ Prime factors of 324 are: 2 x 2 x 3 x 3 x 3

∴ HCF of 252 and 324 = 2 x 2 x 3 x 3

                                        = 36

∴ The greatest number which can divide 257 and 329 so as to leave a remainder of 5 in each case = 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

i hope to this answer