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Master CAT Remainders Questions with practice questions and detailed solutions.
When $10^{100}$ is divided by 7, the remainder is
Text Explanation:
To find the value of $10^{100}mod\left(7\right)$
When 10 is divided by 7, it leaves a remainder 3, so the above equation can be written as,
$3^{100}mod\left(7\right)$
Now looking at the cyclicality of powers of 3 when divided by 7,
$3^1$ mod $7=3$
$3^2$ mod $7=2$
$3^3$ mod $7=6$
$3^4$ mod $7=4$
$3^5$ mod $7=5$
$3^6$ mod $7=1$
From this calculation, it is evident that the powers of 3 modulo 7 repeat every 6 steps. This forms a cycle: 3, 2, 6, 4, 5, 1
$3^{100}=\left(3^6\right)^{16}\times\ \left(3^4\right)$
Since $3^6$ mod $7=1$
We just need to consider $3^4$ mod $7$ which equals 4
Hence the answer is 4.
When $3^{333}$ is divided by 11, the remainder is
There are multiple ways of solving these sorts of questions. One method is to look for powers of the term in the numerator that leave a remainder of 1 or -1 when divided by the denominator.
Noting down the powers of 3, 3, 9, 27, 81, 243
243 is one such number, 242 is multiple of 11 (11 times 22), hence 243 will leave a remainder of 1 when divided by 11.
243 is 3 raised to power 5; we can rewrite the given term as $\dfrac{3^{330}\times\ 3^3}{11}$
The overall remainder will be $\left[\dfrac{3^{330}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$
$\left[\dfrac{3^{5\times\ 66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$
$\left[\dfrac{243^{66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$
$1^{66}\times\ \left[\dfrac{27}{11}\right]_R$
$1\times\ 5$
$5$
Therefore, Option A is the correct answer.
If $10^{68}$ is divided by 13, the remainder is
There are multiple ways of solving such questions involving remainders; one easy way is to look for a power of numerator that leaves a remainder of 1 or -1 when divided by the denominator.
In this instance, 1000, when divided by 13, leaves a remainder of -1
We can rewrite the numerator as $\frac{10^{66}\times\ 100}{13}$
The remainder would be $\left[\frac{10^{66}}{13}\right]_R\times\ \left[\frac{100}{13}\right]_R$
$\left(-1\right)^{22}\times\ 9$
9
Therefore, Option C is the correct answer.
How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?
The number of multiples of 2 between 1 and 120 = 60
The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12
The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7
Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 - 60 - 12 - 7 = 41
