recategorized by
1,186 views
1 votes
1 votes

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

  1. $0$
  2. $1$
  3. $2$
  4. $3$
recategorized by

1 Answer

2 votes
2 votes
Answer: $\mathbf A$
Explanation:
 
$x ≡ 2 \;\text{mod}\; 15$ 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.
 
 
edited by

Related questions

3 votes
3 votes
2 answers
1
Arjun asked Sep 23, 2019
1,644 views
If $a,b$ are positive real variables whose sum is a constant $\lambda$, then the minimum value of $\sqrt{(1+1/a)(1+1/b)}$ is$\lambda \: – 1/\lambda$$\lambda + 2/\lambda...
1 votes
1 votes
1 answer
2
Arjun asked Sep 23, 2019
738 views
Consider the polynomial $x^5+ax^4+bx^3+cx^2+dx+4$ where $a,b,c,d$ are real numbers. If $(1+2i)$ and $(3-2i)$ are two two roots of this polynomial then the value of $a$ i...
1 votes
1 votes
1 answer
3
Arjun asked Sep 23, 2019
594 views
The number of real solutions of the equations $(9/10)^x = -3+x-x^2$ is$2$$0$$1$none of the above
3 votes
3 votes
3 answers
4
Arjun asked Sep 23, 2019
1,370 views
Let $x$ be a positive real number. Then$x^2+\pi ^2 + x^{2 \pi} x \pi+ (\pi + x) x^{\pi}$$x^{\pi}+\pi^x x^{2 \pi} + \pi ^{2x}$$\pi x +(\pi+x)x^{\pi} x^2+\pi ^2 + x^{2 \...