Past JEE PYQs (Integer type)

Hint

(b) Using the Binomial Theorem: The expression is expanded using the binomial theorem for \((1920-1)^{1919}\). All terms except the last two contain a factor of \(1920^2\) (or higher powers), which is a multiple of \(100\left(1920^2=3,686,400\right)\). Any term that is a multiple of 100 does not affect the last two digits of the final number. These terms are represented collectively as \(100 \lambda\).
\(
(1919)^{1919}=100 \lambda+{ }^{1919} \mathrm{C}_{1918}(1920)^1-{ }^{1919} \mathrm{C}_{1919}
\)
Simplifying the Remaining Terms: The expression simplifies to the last two terms:
\(
100 \lambda+1919 \times 1920-1
\)
Calculating the Relevant Value: The value of the non- \(100 \lambda\) part is calculated:
\(
1919 \times 1920-1=3,684,480-1=3,684,479
\)
Finding the Last Two Digits: The full number is therefore \(100 \lambda+3,684,479\). Since \(100 \lambda\) ends in two zeros, the last two digits of the entire expression are determined solely by the last two digits of \(3,684,479\), which are 79.
The Final Product: The text concludes: “Number having last two digit 79 : Product of last two digit 63”. This final step refers to multiplying the two digits themselves:
Product of last two digits \(=7 \times 9=63\)
Explanation: Rewrite the number
\(
1919^{1919}=(1920-1)^{1919}
\)
Apply the binomial theorem
\(
(1920-1)^{1919}=\sum_{k=0}^{1919}\binom{1919}{k}(1920)^{1919-k}(-1)^k
\)
\(
=\binom{1919}{0} 1920^{1919}-\binom{1919}{1} 1920^{1918}+\binom{1919}{2} 1920^{1917}-\cdots-\binom{1919}{1919}
\)
Observe which terms affect the last two digits
Since
\(
1920=100 \times 19+20
\)
any term containing \(1920^2\) or higher powers is divisible by 100.
So these terms do not affect the last two digits.
Hence, only the last two terms matter:
\(
\binom{1919}{1918} 1920-\binom{1919}{1919}
\)
Simplify the remaining terms
\(
\binom{1919}{1918}=1919
\)
So we get:
\(
1919 \cdot 1920-1
\)
Find the last two digits (For example:\(3479=100 \cdot 34+79\), which gives Last two digits = 79)
Reduce modulo 100:
\(
\begin{gathered}
1919 \cdot 1920 \equiv 19 \cdot 20=380 \equiv 80(\bmod 100) \\
80-1=79
\end{gathered}
\)
So, the last two digits of \(1919^{1919}\) are 79.
Find the product of the last two digits
\(
7 \times 9=63
\)

Hint

(a) Step 1: Start with the binomial expansion
\(
(1-x)^{20}=\binom{20}{0}-\binom{20}{1} x+\binom{20}{2} x^2-\binom{20}{3} x^3+\cdots+(-1)^{20}\binom{20}{20} x^{20}
\)
Step 2: Divide by \(\boldsymbol{x}^{\mathbf{2}}\)
\(
\frac{(1-x)^{20}}{x^2}=\frac{\binom{20}{0}}{x^2}-\frac{\binom{20}{1}}{x}+\binom{20}{2}-\binom{20}{3} x+\binom{20}{4} x^2-\cdots+\binom{20}{20} x^{18}
\)
Step 3: Differentiate twice
Differentiate term-by-term:
First derivative
\(
\frac{d}{d x}\left(\frac{(1-x)^{20}}{x^2}\right)=-2\binom{20}{0} x^{-3}+\binom{20}{1} x^{-2}-\binom{20}{3}+2\binom{20}{4} x-\cdots
\)
Second derivative
\(
\frac{d^2}{d x^2}\left(\frac{(1-x)^{20}}{x^2}\right)=6\binom{20}{0} x^{-4}-2\binom{20}{1} x^{-3}+2\binom{20}{4}-6\binom{20}{5} x+\cdots
\)
Step 4: Substitute \(x=1\)
At \(x=1\) :
\(
=6\binom{20}{0}-2\binom{20}{1}+2\binom{20}{4}-6\binom{20}{5}+12\binom{20}{6}-\cdots+18 \cdot 17\binom{20}{20}
\)
This matches exactly the given series:
\(
2 \cdot 1\binom{20}{4}-3 \cdot 2\binom{20}{5}+4 \cdot 3\binom{20}{6}-\cdots+18 \cdot 17\binom{20}{20}
\)
Let this required sum be \(A\).
So we have:
\(
\text { Value at } x=1=6-2\binom{20}{1}+A
\)
Step 5: Evaluate the LHS directly
\(
\frac{(1-x)^{20}}{x^2} \Rightarrow \frac{d^2}{d x^2}\left(\frac{(1-x)^{20}}{x^2}\right)
\)
Substitute \(x=1\) :
\(
(1-1)^{20}=0 \Rightarrow \text { value }=0
\)
Thus,
\(
\begin{gathered}
6-2(20)+A=0 \\
6-40+A=0 \\
A=34
\end{gathered}
\)

Hint

(b) Step 1: Find the value of \(\left(a_1+a_3+a_5+\ldots+a_{19}\right)\)
Let \(P(x)=\left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}\).
Substitute \(x=1\) into the equation:
\(
P(1)=(1+1+1)^{10}=3^{10}=a_0+a_1+a_2+\ldots+a_{20}
\)
Substitute \(x=-1\) into the equation:
\(
P(-1)=\left(1-1+(-1)^2\right)^{10}=(1)^{10}=1=a_0-a_1+a_2-a_3+\ldots+a_{20}
\)
Subtract the second equation from the first to get the sum of odd coefficients:
\(
\begin{gathered}
P(1)-P(-1)=3^{10}-1=2\left(a_1+a_3+a_5+\ldots+a_{19}\right) \\
a_1+a_3+a_5+\ldots+a_{19}=\frac{3^{10}-1}{2}
\end{gathered}
\)
Step 2: Find the value of \(a_2\)
The coefficient of \(x^2, a_2\), can be found by considering the general term of the trinomial expansion or by using the binomial theorem repeatedly.
The coefficient \(a_2\) is 55.
Step 3: Solve for \(\boldsymbol{k}\)
Substitute the values into the given equation:
\(
\begin{gathered}
\left(a_1+a_3+a_5+\ldots+a_{19}\right)-11 a_2=121 k \\
\frac{3^{10}-1}{2}-11 \times 55=121 k
\end{gathered}
\)
\(
\begin{gathered}
\frac{59049-1}{2}-605=121 k \\
\frac{59048}{2}-605=121 k \\
29524-605=121 k \\
28919=121 k \\
k=\frac{28919}{121} \\
k=239
\end{gathered}
\)

Hint

(d) Step 1: Evaluate
\(
\alpha=1+\sum_{r=1}^6(-3)^{r-1}\binom{12}{2 r-1}
\)
Key trick (odd-term binomial identity)
Consider:
\(
(1+x)^{12}-(1-x)^{12}=2\left[\binom{12}{1} x+\binom{12}{3} x^3+\cdots+\binom{12}{11} x^{11}\right]
\)
Divide by \(2 x\) :
\(
\frac{(1+x)^{12}-(1-x)^{12}}{2 x}=\sum_{r=1}^6\binom{12}{2 r-1} x^{2 r-2}
\)
Now substitute \(x^2=3 \Rightarrow x=\sqrt{3}\) :
\(
\sum_{r=1}^6\binom{12}{2 r-1} 3^{r-1}=\frac{(1+\sqrt{3})^{12}-(1-\sqrt{3})^{12}}{2 \sqrt{3}}
\)
But our sum has \((-3)^{r-1}\), so put \(x=i \sqrt{3}\) :
\(
\sum_{r=1}^6\binom{12}{2 r-1}(-3)^{r-1}=\frac{(1+i \sqrt{3})^{12}-(1-i \sqrt{3})^{12}}{2 i \sqrt{3}}
\)
Step 2: Simplify the expression
\(
1+i \sqrt{3}=2 e^{i \pi / 3}
\)
So,
\(
\begin{gathered}
(1+i \sqrt{3})^{12}=2^{12} e^{i 4 \pi}=4096 \\
(1-i \sqrt{3})^{12}=4096
\end{gathered}
\)
Hence,
\(
\sum_{r=1}^6(-3)^{r-1}\binom{12}{2 r-1}=\frac{4096-4096}{2 i \sqrt{3}}=0
\)
Therefore:
\(
\alpha=1
\)
Substitute the value of \(\alpha=1\) into the line equation \(\alpha x-\sqrt{3} y+1=0\) to get:
\(
\alpha x-\sqrt{3} y+1=0 \Rightarrow x-\sqrt{3} y+1=0
\)
Step 3: Calculate the distance
The distance \(\boldsymbol{d}\) from a point ( \(\boldsymbol{x}_{\mathbf{0}}, \boldsymbol{y}_{\mathbf{0}}\) ) to the line \(\boldsymbol{A} \boldsymbol{x}+\boldsymbol{B} \boldsymbol{y}+\boldsymbol{C}=\mathbf{0}\) is given by the formula:
\(
d=\frac{\left|A x_0+B y_0+C\right|}{\sqrt{A^2+B^2}}
\)
For our point \((12, \sqrt{3})\) and line \(x-\sqrt{3} y+1=0(A=1, B=-\sqrt{3}, C=1)\), the distance is:
\(
\begin{gathered}
d=\frac{|1(12)+(-\sqrt{3})(\sqrt{3})+1|}{\sqrt{1^2+(-\sqrt{3})^2}} \\
d=\frac{|12-3+1|}{\sqrt{1+3}} \\
d=\frac{|10|}{\sqrt{4}} \\
d=\frac{10}{2}=5
\end{gathered}
\)
The distance of the point \((12, \sqrt{3})\) from the line \(\alpha x-\sqrt{3} y+1=0\) is 5 units.

Hint

(c) The general term in the expansion of \(\left(1+2^{1 / 3}+3^{1 / 2}\right)^6\) is given by the multinomial theorem as:
\(
T_{p, q, r}=\frac{6!}{p!q!r!}(1)^p\left(2^{1 / 3}\right)^q\left(3^{1 / 2}\right)^r=\frac{6!}{p!q!r!} 2^{q / 3} 3^{r / 2}
\)
where \(p+q+r=6\), and \(p, q, r\) are non-negative integers. For the term to be rational, the exponents of 2 and 3 must be integers. Thus, \(q\) must be a multiple of \(3(q \in\{0,3,6\})\) and \(r\) must be a multiple of \(2(r \in\{0,2,4,6\})\).
Step 2: List all valid combinations of exponents
We list the possible combinations of \((p, q, r)\) that satisfy \(p+q+r=6\) and the rationality conditions:
If \(q=0, r=0 \Longrightarrow p=6\). Combination: \((6,0,0)\).
If \(q=0, r=2 \Longrightarrow p=4\). Combination: \((4,0,2)\).
If \(q=0, r=4 \Longrightarrow p=2\). Combination: \((2,0,4)\).
If \(q=0, r=6 \Longrightarrow p=0\). Combination: \((0,0,6)\).
If \(q=3, r=0 \Longrightarrow p=3\). Combination: \((3,3,0)\).
If \(q=3, r=2 \Longrightarrow p=1\). Combination: \((1,3,2)\).
If \(q=6, r=0 \Longrightarrow p=0\). Combination: \((0,6,0)\). Other combinations violate \(p \geq 0\) or \(p+q+r=6\).
Step 3: Calculate the value of each rational term
We calculate the value for each valid combination:
\((6,0,0): \frac{6!}{6!0!0!} 2^0 3^0=1 \times 1=1\).
\((4,0,2): \frac{6!}{4!0!2!} 2^0 3^1=15 \times 3=45\).
\((2,0,4): \frac{6!}{2!0!4!} 2^0 3^2=15 \times 9=135\).
\((0,0,6): \frac{6!}{0!0!6!} 2^0 3^3=1 \times 27=27\).
\((3,3,0): \frac{6!}{3!3!0!} 2^1 3^0=20 \times 2=40\).
\((1,3,2): \frac{6!}{1!3!2!} 2^1 3^1=60 \times 6=360\).
\((0,6,0): \frac{6!}{0!6!0!} 2^2 3^0=1 \times 4=4\).
Step 4: Sum all the rational terms
The sum of all rational terms is the sum of the values calculated above:
\(
\text { Sum }=1+45+135+27+40+360+4=612
\)

Hint

(a) Given
\(
\sum_{r=1}^{30} \frac{r^2\left(\binom{30}{r}\right)^2}{\binom{30}{r-1}}=\alpha \times 2^{29}
\)
Step 1: Simplify the general term
Use the identity:
\(
\frac{\binom{30}{r}}{\binom{30}{r-1}}=\frac{31-r}{r}
\)
So,
\(
\frac{r^2\left(\binom{30}{r}\right)^2}{\binom{30}{r-1}}=r^2\binom{30}{r} \cdot \frac{31-r}{r}=r(31-r)\binom{30}{r}
\)
Hence the sum becomes:
\(
\sum_{r=1}^{30} r(31-r)\binom{30}{r}
\)
Step 2: Split the sum
\(
r(31-r)=31 r-r^2
\)
So,
\(
\sum_{r=1}^{30} r(31-r)\binom{30}{r}=31 \sum r\binom{30}{r}-\sum r^2\binom{30}{r}
\)
Step 3: Use standard binomial sums
For \(n=30\) :
\(
\begin{gathered}
\sum r\binom{30}{r}=30 \cdot 2^{29} \\
\sum r^2\binom{30}{r}=30 \cdot 31 \cdot 2^{28}
\end{gathered}
\)
Step 4: Substitute
\(
\begin{gathered}
=31\left(30 \cdot 2^{29}\right)-\left(30 \cdot 31 \cdot 2^{28}\right) \\
=30 \cdot 31 \cdot 2^{28}(2-1) \\
=30 \cdot 31 \cdot 2^{28}
\end{gathered}
\)
Rewrite in terms of \(2^{29}\) :
\(
\begin{gathered}
30 \cdot 31 \cdot 2^{28}=\frac{30 \cdot 31}{2} \cdot 2^{29} \\
=465 \cdot 2^{29}
\end{gathered}
\)
Step 5: Identify \(\alpha\)
\(
\alpha=465
\)

Hint

(b) We evaluate
\(
S=\sum_{r=0}^5 \frac{\binom{11}{2 r+1}}{2 r+2}
\)
Step 1: Rewrite the denominator
\(
\frac{1}{2 r+2}=\int_0^1 x^{2 r+1} d x
\)
So,
\(
S=\sum_{r=0}^5\binom{11}{2 r+1} \int_0^1 x^{2 r+1} d x=\int_0^1 \sum_{r=0}^5\binom{11}{2 r+1} x^{2 r+1} d x
\)
Step 2: Recognize the odd-power binomial sum
\(
(1+x)^{11}-(1-x)^{11}=2 \sum_{r=0}^5\binom{11}{2 r+1} x^{2 r+1}
\)
Hence,
\(
\sum_{r=0}^5\binom{11}{2 r+1} x^{2 r+1}=\frac{(1+x)^{11}-(1-x)^{11}}{2}
\)
Step 3: Substitute into the integral
\(
S=\frac{1}{2} \int_0^1\left[(1+x)^{11}-(1-x)^{11}\right] d x
\)
Step 4: Integrate
\(
\begin{gathered}
\int_0^1(1+x)^{11} d x=\left[\frac{(1+x)^{12}}{12}\right]_0^1=\frac{2^{12}-1}{12} \\
\int_0^1(1-x)^{11} d x=\left[-\frac{(1-x)^{12}}{12}\right]_0^1=\frac{1}{12}
\end{gathered}
\)
So,
\(
\begin{gathered}
S=\frac{1}{2}\left(\frac{2^{12}-1}{12}-\frac{1}{12}\right)=\frac{2^{12}-2}{24} \\
S=\frac{4094}{24}=\frac{2047}{12}
\end{gathered}
\)
Step 5: Compute \(m-n\)
\(
\begin{gathered}
m=2047, \quad n=12, \quad \operatorname{gcd}(2047,12)=1 \\
m-n=2047-12=2035
\end{gathered}
\)

Hint

(d) Step 1: Simplify the base modulo 21
First, we find the remainder of the base, 428 , when divided by 21 . We calculate \(428 \div 21:\)
\(
428=20 \times 21+8
\)
This means \(428 \equiv 8(\bmod 21)\).
Step 2: Use the property of modular exponentiation
The problem simplifies to finding the remainder of \(8^{2024}\) when divided by 21. We look for a pattern in the powers of 8 modulo 21:
\(
\begin{gathered}
8^1 \equiv 8 \quad(\bmod 21) \\
8^2 \equiv 64 \equiv 1 \quad(\bmod 21)
\end{gathered}
\)
Since \(8^2 \equiv 1(\bmod 21)\), we can rewrite the exponent 2024 as an even number:
\(
8^{2024}=\left(8^2\right)^{1012}
\)
Substituting the modular equivalence, the expression becomes:
\(
\begin{aligned}
\left(8^2\right)^{1012} & \equiv 1^{1012} \quad(\bmod 21) \\
1^{1012} & \equiv 1 \quad(\bmod 21)
\end{aligned}
\)

Alternate:
We find the remainder of
\(
428^{2024}(\bmod 21)
\)
Step 1: Reduce the base modulo 21
\(
428 \equiv 428-21 \times 20=428-420=8 \quad(\bmod 21)
\)
So,
\(
428^{2024} \equiv 8^{2024}(\bmod 21)
\)
Since
\(
21=3 \times 7
\)
we compute modulo 3 and 7 separately.
Modulo 3
\(
\begin{gathered}
8 \equiv 2 \quad(\bmod 3) \\
2^{2024} \equiv(-1)^{2024} \equiv 1 \quad(\bmod 3)
\end{gathered}
\)
Modulo 7
\(
\begin{gathered}
8 \equiv 1 \quad(\bmod 7) \\
1^{2024} \equiv 1 \quad(\bmod 7)
\end{gathered}
\)
Step 2: Combine results
We need a number \(x[latex] such that:
[latex]
\begin{aligned}
& x \equiv 1 \quad(\bmod 3) \\
& x \equiv 1 \quad(\bmod 7)
\end{aligned}
\)
The common solution is:
\(
x \equiv 1 \quad(\bmod 21)
\)