Question

If h is the hcf of 4052 & 420 & h=4052a+ 420b then find value of x and y

Answer 1

Answer:

h is 4

x = 4052

y = -164

Explanation:

Finding HCF using Euclid's division lemma

If we divide 4052 by 420 we get quotient as 9 and remainder as 272

Thus

$4052=420\times 9+272$   ......... (1)

Now taking numbers 420 and 272 and dividing 420 by 272 we get quotient as 1 and remainder as 148

Thus

$420=272\times 1+148$    ......... (2)        

Now taking numbers 272 and 148 and dividing 272 by 148 we get quotient as 1 and remainder as 124

Thus

$272=148\times 1+124$   ......... (3)

Again taking numbers 148 and 124 and dividing 148 by 124 we get quotient as 1 and remainder as 24

Thus

$148=124\times 1+24$   ......... (4)

Again taking numbers 124 and 24 and dividing 124 by 24 we get quotient as 5 and remainder as 4

Thus

$124=24\times 5+4$   ......... (5)

Since now 24 is completely divisible by 4 therefore the remainder at this step i.e. 4 is the required HCF

Thus HCF of 4052 and 420 is 4

To find the value of x and y our goal will be to start from the step 5 and substituting the values from previous steps so that the numbers can be written as some multiple of 272. Then from the step (1) we can easily substitute 272 in terms of 4052 and 420.

$4=124-5\times24$

$4=124-5\times(148-124)$

$4=6\times (272-148)-5\times 148$

$4=6\times 272-11\times 148$

$4=6\times 272-11\times (420-272)$

$4=17\times 272-11\times 420$

$4=17\times (4052-9\times 420)-11\times 420$   (From Step (1))

$4=17\times 4052-153\times 420-11\times 420$

$4=17\times 4052-164\times 420$

Therefore, $x=4052$ and $y=-164$

Hope this helps.