CAT 2025 Lesson : Number Theory - Past Questions

Observation/Strategy
1) From the options we note that there is a definite value of $x$.
2) We cannot substitute options as we need to find the sum of digits of $x$.
3) For Amitabh to win, when Sashi adds $50$ in his turn, the value goes past $1000$ for the first time.
4) If $x$ is smallest possible, then this exchange between the two happened most number of times.
5) We keep $x$ as a variable and form a table to identify the latest point at which Sashi's value could go past $1000$.

Let the number first picked by Amitabh be $x$. The following table provides the values of the number after Amitabh or Sashi, doubles or increases by $50$ respectively.

Round	Amitabh	Sashi
$1$	$2x$	$2x+50$
$2$	$4x+100$	$4x+150$
$3$	$8x+300$	$8x+350$
$4$	$16x+700$	$16x+750$
$5$	$32x+1500$

Note that in round $5$, even with the smallest value of $x = 1$, number of coins of $1532$ is more than $1000$.

For Amitabh to have won with $x$ being minimum, value of coins for Amitabh in Round $4$ $(16x + 700)$ should have just gone past $1000$

$16x+750 \gt 1000$

⇒ $16x \gt 250$

Smallest value of $x$ that satisfies this is $x = 16$, wherein $16x = 256$

Sum of digits of $x = 1 + 6 = 7$

Answer: (3) $7$

Observation/Strategy
1) $Z = 31!$ which is a very big number. We cannot substitute and check
2) As $Z = 31!$, we can express $X = 31! + 1$ and look at the numbers
3) There should be an easy way to identify factor(s) for each of these numbers. Else, this question cannot be answered in $2$ minutes.

$Z = 31!$ and $X = 31! + 1$

The numbers that we need to test for prime are $(31! + 2)$, $(31! + 3)$, …, $(31! + 30)$ and $(31! + 31)$.

$31!$ is divisible by all numbers from $2$ to $31$.

$\therefore \bm{31! + 2}$ is not prime as $\bm{2}$ divides both these terms. $\bm{31! + 3}$ is not prime as $\bm{3}$ divides both these terms.

This is definitely the case for all numbers till $31$, where $\bm{31! + 31}$ is not prime as it is divisible by $\bm{31}$.

$\therefore$ None of the specified numbers are prime.

Answer: (4) None of the above

Observation/Strategy
1) We cannot substitute options as we need to find the digits in the difference.
2) We need to form equation(s) and simplify them to use the information available in the question.

The initial $3$-digit number is $100 h + 10 t + u$. When the digits are reversed, the number is $100 u + 10 t + h$

Difference between the two = $100 h + 10 t + u - (100 u + 10 t + h)$
$= 99 h - 99 u$
$= 99 \times (h - u)$

As $h$ and $u$ are single-digit numbers, $(h - u)$ is also a single-digit number.

$6$ is the only single-digit number which, when multiplied by $9$, results in a product with $4$ in the units digit.

$\therefore (h - u) = 6$

Difference between the two = $99 \times 6 = 594$

The next two digits from right to left are $\bm{9}$ and $\bm{5}$.

Answer: (2) $9$ and $5$