If 11C12+11C23++11C910=(n/m) with (n,m)=1, then n+m is equal to

Year	2026	2025	2024	Total
Questions	9	17	11	37

Q67 jee_main_2025_29_jan_evening Remainder Problems

The remainder, when

7¹⁰³

is divided by 23, is equal to:

A. $14$ $14$
B. $9$ $9$
C. $17$ $17$
D. $6$ $6$

Solution

Related Formula

Modular arithmetic congruence rules:

A \equiv B m Aⁿ \equiv Bⁿ m

Core Logic

We need to compute $7¹⁰³ 23$ $7^{103} \pmod{23}$. Let's break the exponent down into manageable powers:

7¹⁰³ = 7 \cdot (7²)⁵¹ = 7 \cdot (49)⁵¹

Since $49 \equiv 3 23$ $49 \equiv 3 \pmod{23}$:

7¹⁰³ \equiv 7 \cdot 3⁵¹ 23

Step 1: Simplify Higher Power Components

Break down $3⁵¹$ $3^{51}$ using $3³ = 27 \equiv 4 23$ $3^3 = 27 \equiv 4 \pmod{23}$:

3⁵¹ = (3³)¹⁷ = 27¹⁷ \equiv 4¹⁷ 23

So the expression becomes:

7 \cdot 4¹⁷ = 7 \cdot 4 \cdot (4²)⁸ = 28 \cdot 16⁸

Since $28 \equiv 5 23$ $28 \equiv 5 \pmod{23}$ and $16 \equiv -7 23$ $16 \equiv -7 \pmod{23}$:

\equiv 5 \cdot (-7)⁸ = 5 \cdot 7⁸ 23

Step 2: Final Remainder Evaluation

Now compute $7⁸$ $7^8$ using $7² = 49 \equiv 3 23$ $7^2 = 49 \equiv 3 \pmod{23}$:

7⁸ = (7²)⁴ \equiv 3⁴ = 81 23

Since $81 \equiv 12 23$ $81 \equiv 12 \pmod{23}$:

Expression \equiv 5 \cdot 12 = 60 23

Dividing 60 by 23 ( $23 \times 2 = 46$ $23 \times 2 = 46$) yields a remainder of: $60 - 46 = 14$ $60 - 46 = 14$

Pattern Recognition

Using negative remainders (like $16 \equiv -7$ $16 \equiv -7$) reduces large multiplication outputs instantly during binary power modular loops.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Q jee_main_2025_03_april_morning Rational Terms in Expansion

The sum of all rational terms in the expansion of

(2 + \sqrt(3))⁸

is:

A. 16923
B. 3763
C. 33845
D. 18817

Solution

Related Formula

General term in the binomial expansion of $(a+b)ⁿ$ $(a+b)^n$:

Tᵣ₊₁ = ⁿCᵣ \cdot an-r \cdot b^r

Core Logic

For a term to be rational in $(2+\sqrt(3))⁸$ $(2+\sqrt{3})^8$, the power of $\sqrt(3)$ $\sqrt{3}$ must be an even integer. Thus, $r$ $r$ can only take even values from $0$ $0$ to $8$ $8$

r in 0, 2, 4, 6, 8

Step 1: Calculating individual rational terms

Sum up terms explicitly for all valid values of $r$ $r$ :

Sum = ⁸C₀(2)⁸ + ⁸C₂(2)⁶(\sqrt(3))² + ⁸C₄(2)⁴(\sqrt(3))⁴ + ⁸C₆(2)²(\sqrt(3))⁶ + ⁸C₈(\sqrt(3))⁸

Compute the individual numeric values :

$r=0: 1 \cdot 256 = 256$ $r=0: 1 \cdot 256 = 256$
$r=2: 28 \cdot 64 \cdot 3 = 5376$ $r=2: 28 \cdot 64 \cdot 3 = 5376$
$r=4: 70 \cdot 16 \cdot 9 = 10080$ $r=4: 70 \cdot 16 \cdot 9 = 10080$
$r=6: 28 \cdot 4 \cdot 27 = 3024$ $r=6: 28 \cdot 4 \cdot 27 = 3024$
$r=8: 1 \cdot 1 \cdot 81 = 81$ $r=8: 1 \cdot 1 \cdot 81 = 81$

Total Sum = 256 + 5376 + 10080 + 3024 + 81 = 18817

Pattern Recognition

Notice that the sum of all rational terms can also be viewed as the rational part of the expanded configuration, equivalent to $(2+\sqrt(3))⁸ + (2-\sqrt(3))⁸2$ .

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Q59 jee_main_2025_03_april_morning Binomial Coefficients Series

If

Σr=1⁹( r + 32r)\cdot⁹Cᵣ = α ((3)/(2))⁹ - β

, where

α, β in N

, then

(α + β)²

is equal to:

A. 27
B. 9
C. 81
D. 18

Solution

Related Formula

$Σr=1ⁿ r \cdot ⁿCᵣ x^r = nx(1+x)ⁿ⁻¹$
Binomial Theorem expansion: $Σr=0ⁿ ⁿCᵣ x^r = (1+x)ⁿ$ $\sum_{r=0}^n {^nC_r} x^r = (1+x)^n$

Core Logic

Split the given series summation into two independent parts:

Sum = Σr=1⁹ (r)/(2^r) \cdot ⁹Cᵣ + 3Σr=1⁹ (1)/(2^r) \cdot ⁹Cᵣ

Simplify the first sub-sum using the index relation $r \cdot ⁹Cᵣ = 9 \cdot ⁸Cᵣ₋₁$ $r \cdot {^9C_r} = 9 \cdot {^8C_{r-1}}$:

Σr=1⁹ (9)/(2^r) \cdot ⁸Cᵣ₋₁ = (9)/(2)Σr=1⁹ ⁸Cᵣ₋₁((1)/(2))r-1 = (9)/(2)(1 + (1)/(2))⁸ = (9)/(2)((3)/(2))⁸

Simplify the second sub-sum by including missing index $r=0$ $r=0$:

3[Σr=0⁹ ⁹Cᵣ((1)/(2))^r - 1] = 3[(1 + (1)/(2))⁹ - 1] = 3((3)/(2))⁹ - 3

Step 1: Combining the components

Combine both evaluations to fit into requested representation shape:

Total Sum = (9)/(2)((3)/(2))⁸ + 3((3)/(2))⁹ - 3

Convert the fractional leading term:

(9)/(2)((3)/(2))⁸ = 3 \cdot (3)/(2)((3)/(2))⁸ = 3((3)/(2))⁹

Total Sum = 3((3)/(2))⁹ + 3((3)/(2))⁹ - 3 = 6((3)/(2))⁹ - 3

Matching coefficients gives:

α = 6, β = 3

Evaluate the required squared value:

(α + β)² = (6 + 3)² = 81

Pattern Recognition

Splitting variable factors into standard combinatoric property fractions simplifies coefficient conversions with geometric series denominators.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Q63 jee_main_2025_04_april_evening Properties of Binomial Coefficients

If

1² \cdot ( ¹⁵ C₁ ) + 2² \cdot ( ¹⁵ C₂ ) + 3² \cdot ( ¹⁵ C₃ ) + + 15² \cdot ( ¹⁵ C₁₅ ) = 2m \cdot 3ⁿ \cdot 5k

, where m, n, k ∈ N, then m + n + k is equal to :-

A. $19$ $19$
B. $21$ $21$
C. $18$ $18$
D. $20$ $20$

Solution

Related Formula

The general property linking indices to binomial coefficients is:

r \cdot nr = n \cdot n-1r-1

Core Logic

The given series can be structured using summation notation:

S = Σr=1¹⁵ r² \cdot 15r

Apply the identity $r \cdot 15r = 15 \cdot 14r-1$ to reduce one factor of $r$ $r$:

S = Σr=1¹⁵ r \cdot [ 15 \cdot 14r-1 ] = 15 Σr=1¹⁵ r \cdot 14r-1

Step 1: Splitting the linear term

Rewrite the index variable $r$ $r$ as $(r - 1) + 1$ $(r - 1) + 1$ to align with the binomial lower index:

S = 15 Σr=1¹⁵ ((r - 1) + 1) \cdot 14r-1

S = 15 Σr=1¹⁵ (r - 1) \cdot 14r-1 + 15 Σr=1¹⁵ 14r-1

Applying the property again to the first summation term: $(r-1) 14r-1 = 14 13r-2$ :

S = 15 \cdot 14 Σr=2¹⁵ 13r-2 + 15 Σr=1¹⁵ 14r-1

Step 2: Evaluating the Sums and Prime Factorization

Using the standard total sum of binomial coefficients $Σk=0ⁿ nk = 2ⁿ$ $\sum_{k=0}^{n} \binom{n}{k} = 2^n$:

S = 15 \cdot 14 \cdot 2¹³ + 15 \cdot 2¹⁴

Factor out $15 \cdot 2¹³$ $15 \cdot 2^{13}$ from the expression:

S = 15 \cdot 2¹³ (14 + 2) = 15 \cdot 2¹³ (16) = 15 \cdot 2¹³ \cdot 2⁴

S = 15 \cdot 2¹⁷ = (3¹ \cdot 5¹) \cdot 2¹⁷

Matching this with the given format $2^m \cdot 3ⁿ \cdot 5^k$ $2^m \cdot 3^n \cdot 5^k$, we identify: $m = 17$ $m = 17$, $n = 1$ $n = 1$, and $k = 1$ $k = 1$.

Step 3: Calculating the sum of exponents

Evaluating the targeted summation:

m + n + k = 17 + 1 + 1 = 19

Pattern Recognition

For a series of the type $Σ r² nr$ $\sum r^2 \binom{n}{r}$, remember the standard identity shortcut: $n(n-1)2ⁿ⁻² + n2ⁿ⁻¹$ $n(n-1)2^{n-2} + n2^{n-1}$. Plugging in $n=15$ $n=15$ instantly outputs $15(14)2¹³ + 15(2¹⁴)$ $15(14)2^{13} + 15(2^{14})$, bypasses matching terms manually.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Q54 jee_main_2025_04_april_morning Properties of Binomial Coefficients

For an integer

n \geq 2

, if the arithmetic mean of all coefficients in the binomial expansion of

(x + y)2n - 3

is 16, then the distance of the point

P(2n - 1, n² - 4n)

from the line

x + y = 8

is:

A. $\sqrt(2)$ $\sqrt{2}$
B. $2\sqrt(2)$ $2\sqrt{2}$
C. $5\sqrt(2)$ $5\sqrt{2}$
D. $3\sqrt(2)$ $3\sqrt{2}$

Solution

Related Formula

Sum of binomial coefficients for $(x+y)^m$ $(x+y)^m$ is $2^m$ $2^m$. Total number of terms is $m + 1$ $m + 1$. Perpendicular distance formula from $(x₀, y₀)$ $(x_0, y_0)$ to $ax + by + c = 0$ $ax + by + c = 0$:

d = |ax₀ + by₀ + c|\sqrt(a² + b²)

Core Logic

The power is $m = 2n - 3$ $m = 2n - 3$. Sum of coefficients $= 22n - 3$ $= 2^{2n - 3}$. Total terms $= 2n - 2$ $= 2n - 2$. Given Arithmetic Mean:

22n - 32n - 2 = 16 22n - 32(n - 1) = 16 22n - 4 = 16(n - 1)

Testing integer values, $n = 5$ $n = 5$ satisfies the equation perfectly since $2⁶ = 64$ $2^6 = 64$ and $16(5 - 1) = 64$ $16(5 - 1) = 64$.

Step 1: Determine Coordinates of Point P

Substitute $n = 5$ $n = 5$ into $P(2n - 1, n² - 4n)$ $P(2n - 1, n^2 - 4n)$: $x₀ = 2(5) - 1 = 9$ $x_0 = 2(5) - 1 = 9$ $y₀ = 5² - 4(5) = 5$ $y_0 = 5^2 - 4(5) = 5$ Thus, $P = (9, 5)$ $P = (9, 5)$.

Properties of Binomial Coefficients diagram for Q54 - JEE Main 2025 Morning

Step 2: Calculate Perpendicular Distance

Find the distance from $(9,5)$ $(9,5)$ to the line $x + y - 8 = 0$ $x + y - 8 = 0$:

d = | 9 + 5 - 8\sqrt(1² + 1²)| = 6\sqrt(2) = 3\sqrt(2)

Pattern Recognition

Binomial coefficient logic tightly constrains $n$ $n$ to small integers. Use inspection quickly when polynomial-exponential configurations arise.

Chapter Mix

Class 11 Mathematics: Binomial Theorem Class 11 Mathematics: Straight Lines

If ¹¹C₁2+ ¹¹C₂3+ + ¹¹C₉10=(n)/(m)$\frac{^{11}C_1}{2}+\frac{^{11}C_2}{3}+\dots+\frac{^{11}C_9}{10}=\frac{n}{m}$ with (n,m)=1$\gcd(n,m)=1$, then n+m$n+m$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Expand and Complete the Series

Step 2: Calculate Final Fraction

Pattern Recognition

Chapter Mix

Reference Study Guides

More Binomial Theorem Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Simplify Higher Power Components

Step 2: Final Remainder Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating individual rational terms

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Combining the components

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Splitting the linear term

Step 2: Evaluating the Sums and Prime Factorization

Step 3: Calculating the sum of exponents

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Coordinates of Point P

Step 2: Calculate Perpendicular Distance

Pattern Recognition

Chapter Mix

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If $¹¹C₁2+ ¹¹C₂3+ + ¹¹C₉10=(n)/(m)$ with $(n,m)=1$ $\gcd(n,m)=1$, then $n+m$ $n+m$ is equal to