Find the hcf using euclids division lema of the following
[i] 120 and 80
[ii] 680 and 180
[iii] 200 and 120
[iv] 500 and 800
[v] 600 and 112

Hey mate your answer

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,

Where m is an integer such that m =

Case 2: When a = 3q + 1,

a 3 = (3q +1) 3

a 3 = 27q 3 + 27q 2 + 9q + 1

a 3 = 9(3q 3 + 3q 2 + q) + 1

a 3 = 9m + 1

Where m is an integer such that m = (3q 3 + 3q 2 + q)

Case 3: When a = 3q + 2,

a 3 = (3q +2) 3

a 3 = 27q 3 + 54q 2 + 36q + 8

a 3 = 9(3q 3 + 6q 2 + 4q) + 8

a 3 = 9m + 8

Where m is an integer such that m = (3q 3 + 6q 2 + 4q)

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Step-by-step explanation:

Step-by-step explanation:

Step-by-step explanation:

# here is your answer //

Let take b = 7. (/here b<a. and a , b , q and r are the positive integers)

according to Euclid division lemma

a = bq+r. (0≥r>b)

a = 7q+r (0≥r>7)

so r values are

r >0,1,2,3,4,5,6

1) when r value equals to zero (r=0)

a = 7q+r

a = 7q+0

a = 7q

cube both side

a³ = 343q³

a³ = 7(49q³)

a³ = 7m. (/here m is positive integer and m = 49q³)

2) if value of r is equals to 1 (r=1)

a = 7q+r

a = 7q+1

cube both side

(a)³ = (7q+1)³

a³ = 343q³+1³+3(7q)(1)(7q+1)

a³. = 343q³+21q(7q+1) + 1

a³. = 343q³+ 147q² + 21q + 1

a³ = 7(49q³+21q²+3q)+1

a³ = 7m + 1. (/here m is positive integer and m = 49q³+21q²+3q)

3) if r value equals to 6 ( r = 6)

a = 7q+r

a = 7q+6

cube both side

(a)³ = (7q+6)³

a³ = 343q³+216+3(7q)(6)(7q+6)

a³ = 343q³+126q(7q+6)+216

a³ = 343q³+882q²+756q+216

a³ = 7(49q³+127q²+108q)+216

a³ = 7m+216. (/here m is positive integer and m = 49q³+127q²+108q)

I think 7m + 6 is not a cube of any positive integer

// hope it is helpful #R //☺☺

Step-by-step explanation:

let ' a' be any positive integer and b = 3.

we know, a = bq + r , 0 <  r< b.

now, a = 3q + r , 0<r < 3.

the possibilities of remainder = 0,1 or 2

Case I - a = 3q

a² = 9q² .

= 3 x ( 3q²)

= 3m (where m = 3q²)

Case II - a = 3q +1

a² = ( 3q +1 )²

=  9q² + 6q +1

= 3 (3q² +2q ) + 1

= 3m +1 (where m = 3q² + 2q )

Case III - a = 3q + 2

a² = (3q +2 )²

= 9q² + 12q + 4

= 9q² +12q + 3 + 1

= 3 (3q² + 4q + 1 ) + 1

= 3m + 1 ( where m = 3q² + 4q + 1)

From all the above cases it is clear that square of any positive integer ( as in this case a² ) is either of the form 3m or 3m +1.

take the numbers as 3q, 3q+1 and 3q+2 and cube them such that

x=bq+r where q>0 and 0<r<b

the required answer is 36

yess the will have the same centre o

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