Let S = \(m, n): m, n in \1, 2, 3, dots, 50\ \ . If the number of elements (m, n) in S such that 6^m + 9^n is a multiple of 5 is p and the number of elements (m, n) in S such that m + n is a square of a prime number is q , then p + q is equal to......

Numerical Answer Type:
Enter a numerical value Answer: 1333 to 1333 +4 marks

Solution & Explanation

### Related Formula Modular arithmetic reductions for power cycles: a equiv b pmodm Rightarrow a^k equiv b^k pmodm ### Core Logic Analyze condition p: (6^m + 9^n) is divisible by 5. 6 equiv 1 pmod 5 Rightarrow 6^m equiv 1^m equiv 1 pmod 5. 9 equiv -1 pmod 5 Rightarrow 9^n equiv (-1)^n pmod 5. For the sum to be divisible by 5: 1 + (-1)^n equiv 0 pmod 5 Rightarrow (-1)^n = -1. This implies n must be an ODD integer. Since m in \1, 2, dots, 50\, m can be anything (50 choices). Since n must be odd in \1, dots, 50\, n has 25 choices. p = 50 times 25 = 1250. ### Step 1: Compute q Analyze condition q: (m + n) is the square of a prime number. Max value of m+n = 50+50 = 100. Primes whose squares are leq 100: 2, 3, 5, 7. Their squares are 4, 9, 25, 49. So m+n can be 4, 9, 25, 49. Match List-I with List-II:
m+n=4m+n=9m+n=25m+n=49
No. of ways382448
Explanation for counts: If m+n = S, and m, n geq 1, the number of ways is S-1 (since S leq 50). For S=4: 3 ways. For S=9: 8 ways. For S=25: 24 ways. For S=49: 48 ways. q = 3 + 8 + 24 + 48 = 83. ### Step 2: Final Sum p + q = 1250 + 83 = 1333 ### Pattern Recognition Modular exponentiation immediately shrinks large powers to pm 1. The sum m+n=S where 1 le m,n le N has exactly S-1 solutions if S le N, allowing instant combinatorics tallying without manual counting. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations

Reference Study Guides

Permutations And Combinations

More Permutations and Combinations Previous-Year Questions — Page 3

Q51 jee_main_2025_28_jan_morning Permutations under Restrictions
The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is (1) 4608 (2) 5720 (3) 5719 (4) 4607
  • A. 4608
  • B. 5720
  • C. 5719
  • D. 4607

Solution

### Related Formula For a 5-digit number, total permutations with repetition allowed for n digits is given by: textTotal Cases = d_1 times d_2 times d_3 times d_4 times d_5 ### Core Logic We need 5-digit numbers greater than 50000 using digits \0, 1, 2, 3, 4, 5, 6, 7\ under the restriction d_1 + d_5 le 8. Let's analyze the pairs (d_1, d_5) where d_1 in \5, 6, 7\: Case I: d_1 = 5 Rightarrow d_5 in \0, 1, 2, 3\ (4 options) Case II: d_1 = 6 Rightarrow d_5 in \0, 1, 2\ (3 options) Case III: d_1 = 7 Rightarrow d_5 in \0, 1\ (2 options) Total choices for the first and last digits combined = 4 + 3 + 2 = 9 pairs. ### Step 1: Calculating Intermediate Choices The middle three digits (d_2, d_3, d_4) have no restrictions and can each be chosen from any of the 8 available digits. textNumber of ways = 9 times (8 times 8 times 8) = 4608 ### Step 2: Subtracting Boundary Conditions Since the question specifies numbers strictly greater than 50000, we must check if 50000 is included in our count. For d_1=5 and d_5=0, setting d_2=d_3=d_4=0 gives exactly 50000, which is included in the 4608 count. textTotal numbers = 4608 - 1 = 4607 ### Pattern Recognition Always look carefully at edge constraints like "greater than". Counting the number 50000 explicitly avoids typical off-by-one errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations
Q67 jee_main_2025_28_jan_morning Combinatorial Coefficients and Locus
Let ^nC_r - 1 = 28, ^nC_r = 56 and ^nC_r + 1 = 70. Let A(4cost, 4sint), B(2sint, -2cost) and C(3r - n, r^2 - n - 1) be the vertices of a triangle ABC, where t is a parameter. If (3x - 1)^2 + (3y)^2 = alpha, is the locus of the centroid of triangle ABC, then alpha equals: (1) 20 (2) 8 (3) 6 (4) 18
  • A. 20
  • B. 8
  • C. 6
  • D. 18

Solution

### Related Formula Consecutive combinations ratio property: frac^nC_r-1^nC_r = fracrn-r+1 ### Core Logic Setting up ratios between consecutive given coefficients: frac2856 = frac12 = fracrn-r+1 implies 3r = n + 1 quad dots (1) frac5670 = frac45 = fracr+1n-r implies 9r = 4n - 5 quad dots (2) Solving equations (1) and (2) gives r = 3 and n = 8. ### Step 1: Locating Vertices and Centroid Locus Substituting values for point C gives C(1,0). Let the centroid coordinates be (x,y): 3x = 4cos t + 2sin t + 1 implies 3x - 1 = 4cos t + 2sin t 3y = 4sin t - 2cos t + 0 implies 3y = 4sin t - 2cos t ### Step 2: Squaring and Summing Trig Components Squaring and adding both parametric tracking components eliminates t: (3x - 1)^2 + (3y)^2 = (4cos t + 2sin t)^2 + (4sin t - 2cos t)^2 = 16 + 4 = 20 Thus, alpha = 20. ### Pattern Recognition Symmetric parameter sets of form (Acos t + Bsin t)^2 + (Asin t - Bcos t)^2 collapse instantly into A^2 + B^2 via basic Pythagorean identities. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines Class 11 Maths: Permutations and Combinations
Q74 jee_main_2025_03_april_morning Numbers and Digits Sum Criteria
If the number of seven-digit numbers, such that the sum of their digits is even, is m cdot n cdot 10^n [cite: 696], where m, n in \1, 2, 3, dots, 9\ [cite: 697], then m + n is equal to[cite: 697]:
Numerical Answer. Answer: 14 to 14

Solution

### Related Formula Parity property for numerical spaces: Across any continuous span sequence ending in zero, exactly half of the configuration combinations form an even sum of digits while the rest form odd outputs. ### Core Logic Calculate the total possible combinations of 7-digit numbers first [cite: 1465]: textTotal Numbers = 9 times 10 times 10 times 10 times 10 times 10 times 10 = 9,000,000 [cite: 1465] By using fundamental parity distributions across digit configurations, exactly half of these total options have an even sum of digits [cite: 1467]: textEven Sum Count = frac9,000,0002 = 4,500,000 [cite: 1467] ### Step 1: Extracting factors Express the final numeric amount in requested base-10 exponential format shape [cite: 1468]: 4,500,000 = 45 times 10^5 = 9 cdot 5 cdot 10^5 [cite: 1468] Matching structural parameters [cite: 1469]: m = 9, quad n = 5 [cite: 1469] Both numbers belong to set range \1, 2, dots, 9\ [cite: 697]. Find target summation parameter value [cite: 1470]: m + n = 9 + 5 = 14 [cite: 1470] ### Pattern Recognition The last slot digit completely decides final parity choices. This allows splitting total permutation blocks directly in half without tedious calculations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations
Q74 jee_main_2025_04_april_evening Counting Principles
Let m and n , (m < n) be two 2-digit numbers. Then the total numbers of pairs (m, n) , such that gcd(m, n) = 6 , is
Numerical Answer. Answer: 64 to 64

Solution

### Core Logic Since gcd(m,n) = 6, we can define m = 6a and n = 6b, where a and b are coprime integers (gcd(a,b) = 1). Given that m < n, we must have a < b. Both m and n are two-digit numbers, which means 10 le m, n le 99: 10 le 6a le 99 implies 1.66 le a le 16.5 implies 2 le a le 16 10 le 6b le 99 implies 1.66 le b le 16.5 implies 2 le b le 16 Thus, we need to count all coordinate integer pairs (a,b) satisfying 2 le a < b le 16 such that gcd(a,b) = 1. ### Step 1: Systematic Counting by Fixed Value of 'a' Let's list the valid values for b for each choice of a in the range [2, 16]: - a=2: b in \3, 5, 7, 9, 11, 13, 15\ implies 7 text pairs - a=3: b in \4, 5, 7, 8, 10, 11, 13, 14, 16\ implies 9 text pairs - a=4: b in \5, 7, 9, 11, 13, 15\ implies 6 text pairs - a=5: b in \6, 7, 8, 9, 11, 12, 13, 14, 16\ implies 9 text pairs - a=6: b in \7, 11, 13\ implies 3 text pairs - a=7: b in \8, 9, 10, 11, 12, 13, 15, 16\ implies 8 text pairs - a=8: b in \9, 11, 13, 15\ implies 4 text pairs - a=9: b in \10, 11, 13, 14, 16\ implies 5 text pairs - a=10: b in \11, 13\ implies 2 text pairs - a=11: b in \12, 13, 14, 15, 16\ implies 5 text pairs - a=12: b in \13\ implies 1 text pair - a=13: b in \14, 15, 16\ implies 3 text pairs - a=14: b in \15\ implies 1 text pair - a=15: b in \16\ implies 1 text pair ### Step 2: Final Summation Summing up all valid ordered coordinate tracking entries: textTotal = 7 + 9 + 6 + 9 + 3 + 8 + 4 + 5 + 2 + 5 + 1 + 3 + 1 + 1 = 64 ### Pattern Recognition For modular subset counts, convert your boundary targets to factor conditions directly. Listing terms by prime factors reduces counting errors compared to checking every pair from scratch. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations Class 11 Mathematics: Number Theory
Q65 jee_main_2025_07_april_evening Combinatorial Geometry
Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon P of n sides and q be the number of all quadrilaterals that can be formed by joining the vertices of P. If p + q = 126, then the eccentricity of the ellipse fracx^216 + fracy^2n = 1 is:
  • A. frac34
  • B. frac12
  • C. fracsqrt74
  • D. frac1sqrt2

Solution

### Related Formula The combinations identity for consecutive selection values is: ^nC_r + ^nC_r+1 = ^n+1C_r+1 ### Core Logic Number of triangles from n vertices: p = ^nC_3. Number of quadrilaterals from n vertices: q = ^nC_4. Given algebraic rule: p + q = 126 implies ^nC_3 + ^nC_4 = 126 Applying Pascal's identity: ^n+1C_4 = 126 ### Step 1: Solve for n We need to find n such that ^n+1C_4 = 126: frac(n+1)n(n-1)(n-2)24 = 126 (n+1)n(n-1)(n-2) = 3024 = 9 cdot 8 cdot 7 cdot 6 Equating the consecutive terms: n + 1 = 9 implies n = 8 ### Step 2: Calculate Eccentricity Substitute n = 8 into the ellipse equation: fracx^216 + fracy^28 = 1 Here, a^2 = 16 and b^2 = 8. e = sqrt1 - fracb^2a^2 = sqrt1 - frac816 = sqrtfrac12 = frac1sqrt2 ### Pattern Recognition Pascal's combination identity avoids dealing with tedious polynomial expansions when solving multi-vertex geometry systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations Class 11 Mathematics: Conic Sections

More Permutations and Combinations Questions — jee_main_2026_21_jan_morning

Practice all Permutations and Combinations previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)
YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Now Live Open
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open
Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open
Latest from the Blog
View all →

Loading articles...