- A0
- B1
- C3
- ✓5
⇒ The given number is of the form, 143x + 31
⇒ n = 143x + 31, where x is the quotient
⇒ n = 13(11x) + 13(2) + 5
⇒ n = 13(11x + 2) + 5
So, here the remainder will be 5 when divided by 13
30 questions · timed · auto-graded
$(2^3 \times 3^2 \times 5), (2^2 \times 3^3 \times 5^2)$ and $(2^4 \times ^3\times 5^3 \times 7)$
$HCF = 2^2 \times 3 \times 5 = 60$
A number is a terminating decimal, if the denominator is of the form $2^m \times 5^n$,
where m and n are non$-$negative integers.
$\frac{124}{165}=\frac{124}{3\times5\times11}$
$\frac{131}{30}=\frac{131}{2\times3\times5}$
$\frac{2027}{625}=\frac{2027}{5^4}=\frac{2027}{2^0\times5^4}$
$\frac{1625}{462}=\frac{1625}{2\times3\times7\times11}$
Clearly, option (c) is a terminating decimal, since its denominator is of the form $2^m \times 5^n.$
A number is a terminating decimal, if the denominator is of the form $2^m \times 5^n$,
where m and n are non$-$negative integers.
The prime factorisation of the denominator is $2 \times 3 \times 50^2$
So, the denominator will be non$-$ terminating.
Since $\frac{71}{150}$ is a rational number, it will surely be repeating.
To find the least number divisible by all the natural numbers is the $\text{LCM}$ of the numbers from $1$ to $10$
Find the prime factorization of each of the numbers to find the $\text{LCM}$.
$1, 2, 3, 5, 7, 4 = 2^2, 6 = 2 \times 3, 8 = 2^3, 9 = 3^2, 10 = 2 \times 5$
$\text{LCM} = 2^3 \times 3^2 \times 5 \times 7 = 2520$
The prime factorisation of the denominator is $2 \times 5^2$
Since $4 > 1,$
The decimal expansion will terminate after $4$ decimal places.