Question

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Find hcf of 420 and 272 by division algorithm

Show that any positive odd integer in the form of 5q+1,5q+3​

Answer 1

Step-by-step explanation:

HCF of 420 and 272 is 4

Step-by-step explanation:

Given : Numbers 420 and 272.

To find : Obtain the H.C.F of 420 and 272 by using Euclid 's division algorithm and verify the same by using fundamental theorem of arithmetic ?

Solution :

Applying Euclid 's division algorithm,

H.C.F of 420 and 272

420 = 272\times 1 + 148420=272×1+148

272 = 148\times 1 +124272=148×1+124

148 = 124\times 1 +24148=124×1+24

124 = 24\times 5 + 4124=24×5+4

24 = 4\times 6+ 024=4×6+0

Now, The remainder becomes 0.

So, HCF of 420 and 272 is 4.

Applying fundamental theorem of arithmetic for verifying,

The prime factors are

420=2\times 2\times3\times 5\times7420=2×2×3×5×7

272=2\times 2\times2\times2\times 17272=2×2×2×2×17

HCF(420,272)=2\times 2HCF(420,272)=2×2

HCF(420,272)=4HCF(420,272)=4

Therefore, HCF of 420 and 272 is 4

Answer 2

Here, 420>272

420=272x1+148

272=148×1+124

148=124×1+24 124=24x5+4

24=4x6+0

Since the remainder is O, thus the HCF is 4. The prime factorization of 420 and 272 are:

420=2×2×3×5×7

272=2×2×2×2×17let 'a' be any positive odd integer & a = 5

Then by Euclid's Division Algorithm,

a = bq + r ,

(0 less than or equal to r less than b)

a = 5q + r

(0 less than or equal to r less than 5)

1896

So, r = 0,1,2,3,4

a = 5q +0

a = 5q

Here, 5q is exactly divisible by 2, hence,

it is an even positive number.

a = 5q + 1

Here, 5q + 1 is not exactly divisible by 2,

hence, it is an odd positive number.

.., HCF=2×2=4

2) let 'a' be any positive odd integer & a = 5

Then by Euclid's Division Algorithm,

a = bq + r ,

(0 less than or equal to r less than b)

a = 5q + r

(0 less than or equal to r less than 5)

1896

So, r = 0,1,2,3,4

a = 5q +0

a = 5q

Here, 5q is exactly divisible by 2, hence,

it is an even positive number.

a = 5q + 1

Here, 5q + 1 is not exactly divisible by 2,

hence, it is an odd positive number.

a = 5q + 2

Here, 5q + 2 is exactly divisible by 2,

hence, it is an even positive number.

1896

a = 5q +3

Here, 5q + 3 is not exactly divisible by 2,

hence, it is an odd positive number.

The procedure continues. Hence, any positive integer will be of form 5q + 1, 5q + 3...

where q is some integer.