Real Numbers Test - 1

Explanation:

Dividend = Divisor××Quotient + Remainder

=> Number(dividend)= D x Q + R

Therefore the number (Dividend) = 61××27 + 32

= 1647 + 32

= 1679

Explanation:

Let aa be any positive integer and b=2

Then by applying Euclid’s Division Lemma, we have,

a = 2q + r where 0⩽r <2 r = 0 or 1

Therefore, a=2q or 2q+1

Thus, it is clear that a=2q

i.e., aa is an even integer in the form of 2q

Explanation:

Euclid’s Division Lemma states that for given positive integer aa and bb,

there exist unique integers qq and rr satisfying a = bq + r; 0 ⩽r <b.

Explanation:

Since aa is positive integer,
therefore, rr = 0, 1, 2  only

so that a=3q,3q+1,3q+2.

Explanation:

For the relation x=qy+r, 0 ⩽r <y. 

So, here rr lies between 0 ⩽r <6.

Hence r = 0, 1, 2, 3, 4, 5

Explanation:

x = q y + r

⇒ 27 = 5 × 5 + 2

⇒ q = 5,r = 2

Explanation:

Euclid’s Division Lemma states that

for given positive integer aa and b, there exist unique integers q and r

satisfying a=bq+r;

0 ⩽r <b.

let b=6 then possible values of r will be 0,1,2,3,4,5

when b =6 , r = 0   then  a = 6q +0

                    r = 1            a = 6q +1

                    r = 2            a = 6q +2

                    r = 3            a = 6q + 3

                    r = 4            a = 6q +4

                   r = 5             a = 6q + 5

but 6q , 6q + 2, 6q + 4 cannot be because they are all positive even integers while a is odd integer

thus we can say that a can be 6q + 1 or 6q + 3 or 6q + 5

hence general form is bq + 1

Explanation:

Let nn be a positive integer,

then three consecutive positive integers are (n+1) (n+2) (n+3) = n (n+1) (n+2) + 3 (n+1) (n+2)

Here first term is divisible by 6

and second term is also divisible by 6

because it contains a factor 3 and one of the two consecutive integers (n+1) or (n+2) is even

and thus is divisible by 2

Therefore, the sum of multiple of 6 is also a multiple of 6.