CAT 2025 Lesson : Divisibility - Past Questions

Observation/Strategy
$1$) We are dealing with even integers only. So, we look for numbers divisible by $14$ and $18$ instead of $7$ and $9$.
2) We find the number of $14$ or $18$ multiples and subtract them from the total.
3) Multiples of LCM($14$, $18$) $= \space 126$ will be double counted as they are multiples of both $14$ and $18$.

$100 \le n \le 200$ , where $n$ is even.
Number in the range are from $2 \times 50$ to $2 \times 100$
Total numbers in range $= 100 - 50 + 1 = \bm{51}$

$14$ multiples in the range are from $14 \times 8$ to $14 \times 14$
Number of $\bm{14}$ multiples in the range $= 14 - 8 + 1 = 7$

$18$ multiples in the range are from $18 \times 6$ to $18 \times 11$
Number of $\bm{18}$ multiples in the range $= 11 - 6 + 1 = \bm{6}$

$126$ multiples in the range $= \bm{1}$ (which is $126$ itself)

Number of $14$ or $18$ multiples in range $= 7 + 6 - 1 = \bm{12}$

Numbers that are not multiples of $14$ or $18$ in the range $= 51 - 12 = \bm{39}$

Answer: (3) $39$

Observation/Strategy
$1$) $500 = 2^2 \times 5^3$. $2$ and $5$ are it's only prime factors.
$2$) The base of $(a + b)$ should be the smallest multiple of $2$ and $5$, which is $10$.
$3$) Smallest common multiple of $2$ and $5$ is $10$.

$10^{10}$ is divisible by $500$.

∴$a + b = 10$

Least possible value of $a \times b = 1 \times 9 = 9$

Answer: (2)$9$

Observation/Strategy
$1$) The two terms are being multiplied.
$2$) We can ignore $341^{72}$ as it's units digit will be $1$.
$3$) $7$ has a cyclicity of $4$. The power of $153$ has to be written in $4k$ format.

$U\left((8267)^{153} \times (341)^{72} \right) = U\left((8267)^{153} \right)$

$= U(7^{153})$

$153$ when divided by $4$ yields a remainder of $1$.

$U(7^{153}) = U(7^{1}) = 7$

Answer: (3) $7$