JEE MainMathematics

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Finding the Exponent of 2 in the Constant Term of a Binomial Expansion

Q: What does 2^8 · ell where ell is an odd integer mean mathematically?

It means the highest power of 2 dividing the constant term is exactly 2^8. In other words, when factored into primes, the exponent of 2 is exactly 8.

The number of positive integers k such that the constant term in the binomial expansion of (2 x^3+3/(x^k))^12, x ≠ 0 is 2^8 · ell, where ell is an odd integer, is.

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Step-by-step solutionView

Correct answer

There are 2 positive integers $k$ ( $k = 3$ and $k = 6$ ) for which the constant term contains $2^{8}$ as its highest power of 2.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Binomial expression ${(2 x^{3} + \frac{3}{x^{k}})}^{12}$ with $x \neq 0$ and $k \in Z^{+}$ . The constant term is given by $2^{8} \cdot ℓ$ , where $ℓ$ is an odd integer (i.e., $v_{2} (constant term) = 8$ ).

02Goal

Find the total number of positive integer values of $k$ that satisfy this condition.

03Approach

Write the general term $T_{r + 1} = (\binom{12}{r}) (2 x^{3})^{12 - r} (3 x^{- k})^{r}$ or equivalently $(\binom{12}{r}) (2 x^{3})^{r} (3 x^{- k})^{12 - r}$ . Set the net exponent of $x$ to $0$ to find the integer relation between $r$ and $k$ . Then test each valid integer $r$ to check if the power of $2$ in the constant term is exactly $8$ .

04Execute

Using $T_{r + 1} = (\binom{12}{r}) (2 x^{3})^{r} (3 x^{- k})^{12 - r} = (\binom{12}{r}) 2^{r} 3^{12 - r} x^{3 r - k (12 - r)}$ . For the constant term, the power of $x$ must be zero: $3 r - k (12 - r) = 0 ⟹ k = \frac{3 r}{12 - r}$ . Since $k$ is a positive integer and $0 \leq r \leq 12$ , we must have $0 < r < 12$ . Rewriting: $k = \frac{3 r - 36 + 36}{12 - r} = - 3 + \frac{36}{12 - r}$ . Thus $(12 - r)$ must be a positive divisor of $36$ such that $\frac{36}{12 - r} > 3$ , meaning $12 - r < 12$ . The divisors of $36$ less than $12$ are $12 - r \in {1, 2, 3, 4, 6, 9}$ , which give $r \in {11, 10, 9, 8, 6, 3}$ . For $r = 11$ , $k = 36 / 1 - 3 = 33$ . For $r = 10$ , $k = 15$ . For $r = 9$ , $k = 9$ . For $r = 8$ , $k = 6$ . For $r = 6$ , $k = 3$ . For $r = 3$ , $k = 1$ .

05Execute

The constant term is $C = (\binom{12}{r}) 2^{r} \cdot 3^{12 - r}$ . The exponent of $2$ in $C$ , denoted $v_{2} (C)$ , is $v_{2} ((\binom{12}{r})) + r$ . We need $v_{2} (C) = 8$ . Test the candidate $r$ values: - $r = 3$ : $(\binom{12}{3}) = \frac{12 \times 11 \times 10}{6} = 220 = 2^{2} \times 55 ⟹ v_{2} = 2$ . Then $v_{2} (C) = 2 + 3 = 5 \neq 8$ . - $r = 6$ : $(\binom{12}{6}) = 924 = 2^{2} \times 231 ⟹ v_{2} = 2$ . Then $v_{2} (C) = 2 + 6 = 8$ . This matches! (Gives $k = 3$ ). - $r = 8$ : $(\binom{12}{8}) = (\binom{12}{4}) = 495$ , which is odd $⟹ v_{2} = 0$ . Then $v_{2} (C) = 0 + 8 = 8$ . This matches! (Gives $k = 6$ ). - $r = 9$ : $(\binom{12}{9}) = (\binom{12}{3}) = 220 ⟹ v_{2} = 2$ . Then $v_{2} (C) = 2 + 9 = 11 \neq 8$ . - $r = 10$ : $(\binom{12}{10}) = (\binom{12}{2}) = 66 = 2^{1} \times 33 ⟹ v_{2} = 1$ . Then $v_{2} (C) = 1 + 10 = 11 \neq 8$ . - $r = 11$ : $(\binom{12}{11}) = 12 = 2^{2} \times 3 ⟹ v_{2} = 2$ . Then $v_{2} (C) = 2 + 11 = 13 \neq 8$ . Thus, exactly two values of $k$ work: $k = 3$ and $k = 6$ .

✓Verify

For $k = 3$ , $r = 6$ : term is $(\binom{12}{6}) 2^{6} 3^{6} = 924 \times 64 \times 729 = (4 \times 231) \times 2^{6} \times 729 = 2^{8} \times (231 \times 729)$ , where $231 \times 729$ is odd. Valid. For $k = 6$ , $r = 8$ : term is $(\binom{12}{8}) 2^{8} 3^{4} = 495 \times 2^{8} \times 81 = 2^{8} \times (495 \times 81)$ , where $495 \times 81$ is odd. Valid.

Hints that build this answer step by step

Using the general term $T_{r + 1} = (\binom{12}{r}) (2 x^{3})^{12 - r} {(\frac{3}{x^{k}})}^{r}$ , what condition must $r$ and $k$ satisfy for $T_{r + 1}$ to be independent of $x$ ?
$r (k + 3) = 36$ with $0 \leq r \leq 12$ and $k \in Z^{+}$
Since $k$ is a positive integer ( $k \geq 1$ ), we have $k + 3 \geq 4$ . What are the possible integer values of $r$ that divide 36?
$r \in {1, 2, 3, 4, 6, 9}$
The constant term is given by $C = (\binom{12}{r}) 2^{12 - r} 3^{r}$ . We require the 2-adic valuation $v_{2} (C) = v_{2} ((\binom{12}{r})) + 12 - r = 8$ . For which values of $r \in {1, 2, 3, 4, 6, 9}$ does this hold?
$r = 4$ and $r = 6$
What is the total number of valid positive integers $k$ corresponding to these values of $r$ ?
2

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

What does

2^{8} \cdot ℓ

where

ℓ

is an odd integer mean mathematically?

It means the highest power of 2 dividing the constant term is exactly $2^{8}$ . In other words, when factored into primes, the exponent of 2 is exactly 8.

Answer

Accepted answer: 2

There are 2 positive integers $k$ ( $k = 3$ and $k = 6$ ) for which the constant term contains $2^{8}$ as its highest power of 2.

Why each option works or fails

Step-by-step solution

Given: Binomial expression ${(2 x^{3} + \frac{3}{x^{k}})}^{12}$ with $x \neq 0$ and $k \in Z^{+}$ . The constant term is given by $2^{8} \cdot ℓ$ , where $ℓ$ is an odd integer (i.e., $v_{2} (constant term) = 8$ ).
Goal: Find the total number of positive integer values of $T_{r + 1}$ that satisfy this condition.
Approach: Write the general term $T_{r + 1} = (\binom{12}{r}) (2 x^{3})^{12 - r} (3 x^{- k})^{r}$ or equivalently $(\binom{12}{r}) (2 x^{3})^{r} (3 x^{- k})^{12 - r}$ . Set the net exponent of $x$ to $0$ to find the integer relation between $r$ and $k$ . Then test each valid integer $r$ to check if the power of $2$ in the constant term is exactly $8$ .
Execute: Using $T_{r + 1} = (\binom{12}{r}) (2 x^{3})^{r} (3 x^{- k})^{12 - r} = (\binom{12}{r}) 2^{r} 3^{12 - r} x^{3 r - k (12 - r)}$ . For the constant term, the power of $x$ must be zero: $3 r - k (12 - r) = 0 ⟹ k = \frac{3 r}{12 - r}$ . Since $ℓ$ is a positive integer and $v_{2} (constant term) = 8$ , we must have $0 < r < 12$ . Rewriting: $k = \frac{3 r - 36 + 36}{12 - r} = - 3 + \frac{36}{12 - r}$ . Thus $(12 - r)$ must be a positive divisor of $36$ such that $\frac{36}{12 - r} > 3$ , meaning $12 - r < 12$ . The divisors of $36$ less than $12$ are $12 - r \in {1, 2, 3, 4, 6, 9}$ , which give $r \in {11, 10, 9, 8, 6, 3}$ . For $r = 11$ , $k = 36 / 1 - 3 = 33$ . For $r = 10$ , $k = 15$ . For $r = 9$ , $k = 9$ . For $r = 8$ , $k = 6$ . For $r = 6$ , $k = 3$ . For $r = 3$ , $k = 1$ .
Execute: The constant term is $C = (\binom{12}{r}) 2^{r} \cdot 3^{12 - r}$ . The exponent of $2$ in $C$ , denoted $v_{2} (C)$ , is $v_{2} ((\binom{12}{r})) + r$ . We need $v_{2} (C) = 8$ . Test the candidate $r$ values: - $r = 3$ : $(\binom{12}{3}) = \frac{12 \times 11 \times 10}{6} = 220 = 2^{2} \times 55 ⟹ v_{2} = 2$ . Then $v_{2} (C) = 2 + 3 = 5 \neq 8$ . - $r = 6$ : $(\binom{12}{6}) = 924 = 2^{2} \times 231 ⟹ v_{2} = 2$ . Then $v_{2} (C) = 2 + 6 = 8$ . This matches! (Gives $k = 3$ ). - $r = 8$ : $(\binom{12}{8}) = (\binom{12}{4}) = 495$ , which is odd $⟹ v_{2} = 0$ . Then $v_{2} (C) = 0 + 8 = 8$ . This matches! (Gives $k = 6$ ). - $r = 9$ : $(\binom{12}{9}) = (\binom{12}{3}) = 220 ⟹ v_{2} = 2$ . Then $v_{2} (C) = 2 + 9 = 11 \neq 8$ . - $r = 10$ : $(\binom{12}{10}) = (\binom{12}{2}) = 66 = 2^{1} \times 33 ⟹ v_{2} = 1$ . Then $v_{2} (C) = 1 + 10 = 11 \neq 8$ . - $r = 11$ : $v_{2} (C) = 2 + 6 = 8$ . Then $v_{2} (C) = 2 + 11 = 13 \neq 8$ . Thus, exactly two values of $k$ work: $k = 3$ and $k = 6$ .
Verify: For $k = 3$ , $r = 6$ : term is $(\binom{12}{6}) 2^{6} 3^{6} = 924 \times 64 \times 729 = (4 \times 231) \times 2^{6} \times 729 = 2^{8} \times (231 \times 729)$ , where $231 \times 729$ is odd. Valid. For $k = 6$ , $r = 8$ : term is $(\binom{12}{8}) 2^{8} 3^{4} = 495 \times 2^{8} \times 81 = 2^{8} \times (495 \times 81)$ , where $495 \times 81$ is odd. Valid.

Finding the Exponent of 2 in the Constant Term of a Binomial Expansion

What feels right?

There are 2 positive integers $k$ ( $k = 3$ and $k = 6$ ) for which the constant term contains $2^{8}$ as its highest power of 2.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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