Question

25. If the HCF of 408 and 1032 is expressible in the form 1032x2+408xp, then the value
of p is ​

Answer 1

Answer:

Step-by-step explanation:

Firstly we will find the HCF of the given two numbers. So, we know when the remainder becomes zero then the divisor is the HCF of the numbers. Therefore, the HCF of the numbers 408 and 1032 is 24.

HCF of 408 and 1032 is 24(by Euclid division lemma)

ATQ,

HCF = 1032×2+408×P

24=1032×2+408×P

24=2064+408×P

24=2064+408P

-408P=2064-24

-408P=2040

408P=-2040

P=-2040/408

P=-5

Verification

R.H.S

1032×2+408×P

=2064+408(-5)

=2064+(-2040)

=2064-2040

=24=L.H.S