Question

If d=HCF(a,b) , then d=na+mb where m and n are integers. Verify this result for a=384 and b=26. I asked this question last time but made a mistake in symbol.

Answer 1

Using Euclid's division algorithm for 384 and 26, we have
here, 384>26;
∴, 384=26×14+20 --------------(1)
26=20×1+6 ----------------------(2)
20=6×3+2 -----------------------(3)
6=2×3+0
since the remainder is 0, then d=HCF(a,b)=HCF(384,26)=2
Now, from (3) we get,
2=20-(6×3)
or, 2=20-{26-(20×1)}×6 ; [using (2)]
or, 2=26×6+20×(1+1×6)
or, 2=26×6+20×7
or, 2=26×6+{384-(26×14)}×7 ; [using (1)]
or, 2=384×7+26×(6-14×7)
or, 2=384×7+26×(6-98)
or, 2=384×7+26×(-92)
∴, d=na+mb=7×384+(-92)×26
i.e., n=7, m=-92

Answer 2

Answer:

n=7, m=-92

Step-by-step explanation:

Using Euclid's division algorithm for 384 and 26, we have

here, 384>26;

∴, 384=26×14+20 --------------(1)

26=20×1+6 ----------------------(2)

20=6×3+2 -----------------------(3)

6=2×3+0

since the remainder is 0, then d=HCF(a,b)=HCF(384,26)=2

Now, from (3) we get,

2=20-(6×3)

or, 2=20-{26-(20×1)}×6 ; [using (2)]

or, 2=26×6+20×(1+1×6)

or, 2=26×6+20×7

or, 2=26×6+{384-(26×14)}×7 ; [using (1)]

or, 2=384×7+26×(6-14×7)

or, 2=384×7+26×(6-98)

or, 2=384×7+26×(-92)

∴, d=na+mb=7×384+(-92)×26

i.e., n=7, m=-92