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Using the theory that any positive odd integers are of form $4 q + 1$ or $4 q + 3$ where $q$ is a positive integer. If the quotient is $4$ , the dividend is $19$ then what will be the remainder?

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17

2

1

3

answer is D.

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Detailed Solution

Given, that any positive odd integers are of form

4 q + 1

4 q + 3

.
We have to find the remainder of a number whose quotient is 4 and dividend is 19.
Let

a

be any positive integer.
We know by Euclid's algorithm, that if

a

and

b

are two positive integers, there exist unique integers

q

and

r

satisfying,

a = bq + r

where

0 \leq r < b

.
Take

b = 4

we get,

a = 4 q + r, 0 \leq r < b

.
Since

0 \leq r < 4

, therefore the possible remainders are

0, 1, 2

and

3

.
So,

a

can be

4 q, 4 q + 1

, or

4 q + 2

4 q + 3

, where

q

is the quotient.
Since

a

is odd,

a

cannot be

4 q

4 q + 2

as they are both divisible by

2

.
Therefore, any odd integer is of form

4 q + 1

4 q + 3

.
Putting the value of

q = 4

as given, in the odd integers,

4 q + 1 = 4 \times 4 + 1

\Rightarrow 4 q + 1 = 17

Or,

4 q + 3 = 4 \times 4 + 3

\Rightarrow 4 q + 3 = 19

Since the dividend is

19

, the remainder is

3 .

Therefore, the remainder is 3.
Hence, option 4 is correct.

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