Concepts of Indices | HARD | CAT 2025 Slot 1
For any natural number k, let aₖ = 3ᵏ. The smallest natural number m for which (a₁)¹ × (a₂)² × … × (a₂₀)²⁰ < (a₂₁ × a₂₂ × … × a₂₀₊ₘ)⁵ is:
1. 56
2. 59
3. 58
4. 57
Answer
Given: aₖ = 3ᵏ.
Then (aᵢ)ⁱ = (3ᵢ)ⁱ = 3⁽ⁱ²⁾.
Left side:
L = (a₁)¹ × (a₂)² × … × (a₂₀)²⁰ = 3^(1² + 2² + … + 20²)
Sum of squares 1² + 2² + … + 20² = 2870 → So L = 3²⁸⁷⁰.
Right side inside product:
P = a₂₁ × a₂₂ × … × a₂₀₊ₘ = 3²¹ × 3²² × … × 3^(20+m) = 3^(sum of integers from 21 to 20+m).
Raise to power 5: R = P⁵ = 3^(5 × sum of integers from 21 to 20+m)
We need → 3²⁸⁷⁰ < 3^(5 × sum)
So: 2870 < 5 × (sum of 21 to 20+m)
Let N = 20 + m.
Sum from 21 to N = (N(N+1))/2 − 210.
So: 2870 < 5 × [(N(N+1))/2 − 210]
Divide and simplify:
(N(N+1))/2 > 784 → N(N+1) > 1568
Check values:
39×40 = 1560 (too small)
40×41 = 1640 (works)
So the smallest N = 40.
Thus:
m = N − 20 = 40 − 20 = 20 → Final Answer: m = 20
PnC + Factors + Maxima Minima | HARD | CAT 2025 Slot 1
In a 3-digit number N, the digits are non-zero and distinct such that none of the digits is a square, and only one of the digits is a prime number. Then, the number of factors of the minimum possible value of N is ______
Answer
Square digits forbidden: 1, 4, 9. Allowed nonzero digits: 2,3,5,6,7,8.
Exactly one digit must be prime, so the other two must be composite. Allowed composites here are 6 and 8, so the three digits must be {prime, 6, 8}.
To minimize the 3-digit number, put the smallest available digit in the hundreds place. The smallest prime available is 2.
So hundreds = 2, and place the remaining digits in increasing order: tens = 6, units = 8. Minimum N = 268.
Factorize 268 = 2^2 × 67.
Number of factors = (2 + 1) × (1 + 1) = 3 × 2 = 6.
Answer: 6.
Vedic Patterns G Strategy | Inequality | EASY | CAT 2025 Slot 1
The number of distinct pairs of integers (x, y) satisfying the inequalities x > y ≥ 3 and x + y < 14 is ____.
Answer
We fix y and find all possible x values that satisfy both conditions:
x > y
x + y < 14
| y | Allowed x values | Count |
| 3 | 4, 5, 6, 7, 8, 9, 10 | 7 |
| 4 | 5, 6, 7, 8, 9 | 5 |
| 5 | 6, 7, 8 | 3 |
| 6 | 7 | 1 |
| Total | 16 |
Vedic Pattern + Concepts of Logs DPAC G Strategy | EASY | CAT 2025 Slot 1
The number of distinct integers n for which log(1/4) of (n² − 7n + 11) > 0 is _____.
Answer
log(1/4) of (n² − 7n + 11) > 0 is _____.
Using DPAC Strategy: Cycle
n² − 7n + 11 < (1/4)0 → n² − 7n + 11 < 1
Now we need values of n which satisfy 0 < n² − 7n + 11 < 1
Compute E(n) = n² − 7n + 11 for nearby integers:
n = 1 → E = 5 (not in (0,1))
n = 2 → E = 1 (equals 1, excluded)
n = 3 → E = −1 (negative, excluded)
n = 4 → E = −1 (negative, excluded)
n = 5 → E = 1 (equals 1, excluded)
For n ≤ 0 or n ≥ 6, E ≥ 5 (too large).
Therefore no integer n satisfies 0 < E < 1. Final answer: 0.
Note: Official OG says answer is 2. Which is challenged by Cetking
PnC G Strategy | Concepts of PnC | EASY | CAT 2025 Slot 1
A cafeteria offers 5 types of sandwiches. Moreover, for each type of sandwich, a customer can choose one of 4 breads and opt for either small or large sized sandwich. Optionally, the customer may also add up to 2 out of 6 available sauces. The number of different ways in which an order can be placed for a sandwich, is
- 880
- 600
- 800
- 840
Answer
Given 5 sandwiches
4 breads
2 sizes
Sauces: choose 0, 1, or 2 from 6
Number of sauce choices:
0 sauces → 1 way
1 sauce → 6 ways
2 sauces → 15 ways
Total = 22 ways.
Total combinations =
5 × 4 × 2 × 22 = 880.
Answer: 880
