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Consider the following system of equivalences of integers,

$$x \equiv 2 \text{ mod } 15$$

$$x \equiv 4 \text{ mod } 21$$

The number of solutions in $x$, where $1 \leq x \leq 315$, to the above system of equivalences is

- $0$
- $1$
- $2$
- $3$

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$x ≡ 2 \;\text{mod}\; 15$ means $ x-2=15\times t_1 \Rightarrow x=2+15\times t_1$

$x ≡ 4 \;\text{mod}\;21$ means $x-4 = 21 \times t_2 \Rightarrow x=4+ 21\times t_2$

where $t_1$ and $t_2$ are two constants.

For finding the common value of $x$ in both cases, both these equations can be equated with each other.

$2+15\times t_1=4+ 21\times t_2 \Rightarrow 15\times t_1-21\times t_2 = 2$

Hence, no combination of $t_1$ and $t_2$ can satisfy this equation.

So, the number of values of $x$ is 0.

$\therefore \mathbf A$ is the correct option.

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