# Find the value of $x$ that satisfying $150x \equiv 35(Mod31)$ is

$\

A)14 \\

B)22 \\

C)24 \\

D)12 \\

\ $

Answer

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Hint: To proceed with this solution we have to compare the given expression with the standard form.

Given expression is $150x \equiv 35(Mod31)$

Let us compare the expression with standard form i.e. $a \equiv b(\bmod n)$

Before comparing it lets us know what does the standard form means

Here,

$a \equiv b(\bmod n)$ Means $a - b$ is divisible by $n$

So, here if we compare we can say $a = 150x,b = 35$ and $n = 31$

So,

$150x \equiv 35(Mod31)$

$ \Rightarrow 150x - 35$ is divisible by $31$

$ \Rightarrow 5(30x - 7)$ is divisible by $31$

Now here let us say that $30x - 7 = 31k$ where $k$is any natural number.

From this we can equate x value as

$ \Rightarrow $ $x = \dfrac{{31k + 7}}{{30}}$

$ \Rightarrow \dfrac{{30k + (k + 7)}}{{30}}$

$ \Rightarrow \dfrac{{30k}}{{30}} + \dfrac{{k + 7}}{{30}}$

$ \Rightarrow k + \dfrac{{k + 7}}{{30}}$

Here $k + 7$ is divisible by $30$ where $k = 23$ which is smallest possible value

So, if $k = 23$ then

$ \Rightarrow k + \dfrac{{k + 7}}{{30}}$=$k + \dfrac{{23 + 7}}{{30}} = k + \dfrac{{30}}{{30}} = k + 1$

And we know that $k = 23$

So, here

$\

\Rightarrow x = K + 1 \\

\Rightarrow x = 23 + 1 \\

\Rightarrow x = 24 \\

\ $

From this we can say that for $x = 24$ is the value that satisfying the expression

Option C is the correct

NOTE: Problems like the above model have to be compared with standard forms and in the above problem we have to substitute k value in two places in the same equation. Instead of substituting K value in the two places at a time it is better to substitute in one place then make the equation simple so get the answer directly.

Given expression is $150x \equiv 35(Mod31)$

Let us compare the expression with standard form i.e. $a \equiv b(\bmod n)$

Before comparing it lets us know what does the standard form means

Here,

$a \equiv b(\bmod n)$ Means $a - b$ is divisible by $n$

So, here if we compare we can say $a = 150x,b = 35$ and $n = 31$

So,

$150x \equiv 35(Mod31)$

$ \Rightarrow 150x - 35$ is divisible by $31$

$ \Rightarrow 5(30x - 7)$ is divisible by $31$

Now here let us say that $30x - 7 = 31k$ where $k$is any natural number.

From this we can equate x value as

$ \Rightarrow $ $x = \dfrac{{31k + 7}}{{30}}$

$ \Rightarrow \dfrac{{30k + (k + 7)}}{{30}}$

$ \Rightarrow \dfrac{{30k}}{{30}} + \dfrac{{k + 7}}{{30}}$

$ \Rightarrow k + \dfrac{{k + 7}}{{30}}$

Here $k + 7$ is divisible by $30$ where $k = 23$ which is smallest possible value

So, if $k = 23$ then

$ \Rightarrow k + \dfrac{{k + 7}}{{30}}$=$k + \dfrac{{23 + 7}}{{30}} = k + \dfrac{{30}}{{30}} = k + 1$

And we know that $k = 23$

So, here

$\

\Rightarrow x = K + 1 \\

\Rightarrow x = 23 + 1 \\

\Rightarrow x = 24 \\

\ $

From this we can say that for $x = 24$ is the value that satisfying the expression

Option C is the correct

NOTE: Problems like the above model have to be compared with standard forms and in the above problem we have to substitute k value in two places in the same equation. Instead of substituting K value in the two places at a time it is better to substitute in one place then make the equation simple so get the answer directly.

Last updated date: 18th Sep 2023

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