Question

Express the HCF of 148 and 128 as 148x + 128y, where x and y are integers.​

Answer 1

Answer:

Euclidian algorithm gives both the gcd and the integer linear combination.

468=2⋅222+24

222=9⋅24+6

24=4⋅6+0.

Therefore, gcd(468,222)=6 .

Now do the operations in reversed order,

6=222−9⋅24

=222−9⋅(468−2⋅222)

=19⋅222−9⋅468.

The question ask for two different ways to write it. We can give infinite of them. Notice that the least common multiple of 222 and 468 is

222⋅4686=17316,

therefore

6=19⋅222−9⋅468

=19⋅222−9⋅468+17316k−17316k

=19⋅222−9⋅468+78⋅222k−468⋅37k

=(19+78k)⋅222−(9+37k)⋅468.

For example, if k=1 we have

6=97⋅222–46⋅468,

and a new solution for each integer value of k .

Step-by-step explanation:

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Answer 2

Concept

The factor that two or more given integers have in common is known as the highest common factor (HCF). Additionally known as the "Greatest Common Divisor," it (GCD). Because 12 is the biggest number that can completely divide both 24 and 36, for instance, the HCF of these two numbers is 12.

Given

numbers = 148 and 128

expressed as linear equation = 148x + 128y

where x and y are integers.

Find

we are asked to express the HCF ofgiven numbers as the equation

148x + 128y

Solution

so here we have two integers : 148 and 128

Then, by applying Euclid's division lemma, we get

148 = 128 × 1 ₊ 20 .....(1)

since the remainder ≠ 0, so apply division lemma on divisor 128 and remainder 20.

128 = 20 × 6 ₊ 8 .......(2)

since the remainder ≠ 0, so apply division lemma on divisor 20 and remainder 8.

20 = 8 × 2 ₊ 4 .........(3)

since the remainder ≠ 0, so apply division lemma on divisor 8 and remainder 4.

8 = 4 × 2 ₊ 0 ..........(4)

we observe that remainder is 0.

so the last divisor 4 is the HCF of 148 and 128.

Now in order to express the HCF as a linear combination of 148 and 128, we perform

4 = 128 ₋ 20 × 6 [from (2)]

= 128 ₋ (148 ₋ 128 × 1) × 6 [from (1)]

= 128 ₋ (148 × 6) ₊ (128 × 6)

taking 128 common

4 = 128[1₊6] ₋ 148 × 6

4 = 128×7 ₋ 148× 6

4 = 148 (₋6) ₊ 128(7)

let say x = ₋6 and y = 7

then HCF = 148 × (x) ₊ 128 × (y)

hence x = -6 and y = 7

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