Question

for any positive interger a&b,it is known that a=bq+r,where q is some interger and 0<r< b. so if a =17 b=7then r can take value 1) only 3 2) any positive interger less then 7 3) only value greater then and equal to 3 and less then 7 4) zero​

Answer 1

Hello Friend

First of all no one answered your question because of the way you have posted the question.Thank god there is someone who understood your question.

Here is the answer-
A=17
B=7
a=bq+r
We know that
17=7x2+3
=7x0+17
=7x1+10
Since the value of a or r is not given, we cannot conclude what is the exact number, but we can tell that r= 3,10,17

From this you can select the correct option

Answer 2

For any positive integer $a\ \&\ b$ we have $a=bq+r$ where $q$ is some integer and $0\leq r<b.$

From this expression $r$ is given by,

$\longrightarrow r=a-bq\quad\quad\dots(1)$

We're given $a=17$ and $b=7$ and we've to find possible values of $r.$

As per the definition we have $0\leq r<7$ since $b=7.$

Putting values of $a$ and $b$ in (1),

$\longrightarrow r=17-7q\quad\quad\dots(2)$

Thus,

$\longrightarrow 0\leq r<7$

From (2),

$\longrightarrow 0\leq17-7q<7$

Multiplying each by -1, (note the sign change)

$\longrightarrow 0\geq7q-17>-7$

Or, by changing the limits,

$\longrightarrow -7\leq7q-17<0$

Adding 17 to each,

$\longrightarrow 10\leq7q<17$

Dividing each by 7,

$\longrightarrow \dfrac{10}{7}\leq q<\dfrac{17}{7}$

Since $q$ is an integer,

$\Longrightarrow q=2$

That is, $q$ can be 2 only.

Then (2) becomes,

$\longrightarrow r=17-7\times2$

$\longrightarrow r=17-14$

$\longrightarrow\underline{\underline{r=3}}$

Therefore $r$ can take the value of 3 only.

Hence (1) is the answer.