7. The largest number of four digits exactly divisible by 77 is
a. 9768
b. 9933
c. 9988
d. 9944
Answer: B
Explanation:
Find the remainder when 10000 is divided by 77. Then subtract that remainder from 10000.
$\require{enclose}
\begin{array}{rll}
129\,\,\, &\\[2px]
77 \enclose{longdiv}{10000}\kern-.2ex \\[2px]
\underline{77}{\phantom{00}} \\[2px]
230\phantom{0} \\[2px]
\underline {154}{\phantom{0}} \\[2px]
760 \\[2px]
\underline{693} \\[2px]
67
\end{array}$
So largest four digit number divisible by 77 is 10000 – 67 = 9933
8. The number of distinct prime factors of ${2^{2}} \times {3^{5}} \times {5^{4}}$
a. 3
b. 11
c. 90
d. 5
Answer: A
Explanation:
Number of distinct prime factors of ${2^{2}} \times {3^{5}} \times {5^{4}}$ are $2, 3, 5$
9. The general format of a prime number $k$ greater than equal to 5 is
a. $5k + 1$
b. $6k + 1$
c. $6k - 1$
d. Both $B$ and $C$
Answer: D
Explanation:
Any prime greater than or equal to $5$ is in the format of $6k \pm 1$
10. The remainder when square of a prime number greater than equal to 5 divided by 6 is
a. 4
b. 1
c. 5
d. Both $B$ and $C$
Answer: B
Explanation:
Square
We know that prime numbers greater than equal to 5 are in the format of $6k \pm 1$
$\therefore$ ${\left( {6k \pm 1} \right)^2} = 36{k^2} \pm 12k + 1$ $ = 6\left( {6{k^2} \pm 2} \right) + 1$
When the above expression is divided by $6$, we get $1$ as remainder.
11. The number of distinct prime factors of $9!$
a. 9
b. 4
c. 80
d. 160
12. The number of even numbers and 3 multiples from 20 to 90 (both inclusive) are respectively
a. 35, 24
b. 36, 24
c. 35, 23
d. 36, 23
Answer: B
Explanation:
Number of terms in an Arithmetic Progression (A.P) = $n = \dfrac{{l - a}}{d} + 1$ $a$ = First term $l$ = Last term $n$ = Number of terms $d$ = Difference between consecutive terms.
Step1: Let us calculate the number of even numbers. First and last even numbers from 20 to 90 are $20$ and $90$. The formula is $\begin{align}
n &= \dfrac{{l - a}}{d} + 1 \\[4px]
& = \dfrac{{90 - 20}}{2} + 1 \\[4px]
& = 36
\end{align}$
Step2: Now, Let us calculate the number of $3$ multiples. First and last 3 multiples from 20 to 90 are $21$ and $90$. The formula is $\begin{align}
n &= \dfrac{{l - a}}{d} + 1 \\[4px]
& = \dfrac{{90 - 21}}{3} + 1 \\[4px]
& = 24
\end{align}$
