JEE Main 2021 (26 February Shift 2)

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Animated Solution for Mathematics - Permutations and Combinations: A natural number has prime factorization given by $n = 2^{x} 3^{y} 5^{z}$ , where $y$ and $z$ are such that $y + z = 5$ and $y^{- 1} + z^{- 1} = \frac{5}{6}, y > z$ . Then the number of odd divisors of $n$ , including 1, is :

Q: Wait, why do we ignore the $2^x$ term when counting odd divisors?

An odd number, by definition, cannot have any factor of $2$. If a divisor contains a factor of $2$, it becomes an even number. Therefore, to count only the odd divisors of $n = 2^x \cdot 3^y \cdot 5^z$, we must completely exclude the $2^x$ term and focus only on the odd prime factors $3^y$ and $5^z$.

Q: Is there a faster way to find $y$ and $z$ without the quadratic equation?

While you might be able to guess the values $2$ and $3$ by inspection because they are small integers, the quadratic method is the most robust and systematic approach. In JEE Advanced, relying on intuition is risky; having a rigorous algebraic method ensures you never miss a solution or get confused when the numbers are larger or non-integers.

Select Answer:

Visualized Solution

$n = 2^{x} 3^{y} 5^{z}$

Given prime factorization: $n = 2^{x} 3^{y} 5^{z}$
Goal: Find the number of odd divisors of $n$
We need to find the values of $y$ and $z$ first.

$y + z = 5$ and $\frac{1}{y} + \frac{1}{z} = \frac{5}{6}$

Constraint 1: $y + z = 5$
Constraint 2: $\frac{1}{y} + \frac{1}{z} = \frac{5}{6}$
Condition: $y > z$

$\frac{y + z}{y z} = \frac{5}{6}$

Let's simplify the second constraint.
Take the common denominator for $\frac{1}{y} + \frac{1}{z}$ .
We get: $\frac{y + z}{y z} = \frac{5}{6}$

$\frac{5}{y z} = \frac{5}{6}$

Substitute $y + z = 5$ into our simplified equation.
$\frac{5}{y z} = \frac{5}{6}$
Notice the numerators are identical.

$y z = 6$

From $\frac{5}{y z} = \frac{5}{6}$ , we can cancel $5$ from both sides.
This leaves us with $\frac{1}{y z} = \frac{1}{6}$ .
Therefore, $y z = 6$ .

$t^{2} - 5 t + 6 = 0$

We have Sum: $y + z = 5$ and Product: $y z = 6$ .
$y$ and $z$ are roots of a quadratic equation in variable $t$ .
Equation: $t^{2} - (Sum) t + (Product) = 0$
$t^{2} - 5 t + 6 = 0$

$y = 3, z = 2$

Factorize $t^{2} - 5 t + 6 = 0$ :
$(t - 2) (t - 3) = 0$
Roots are $t = 2$ and $t = 3$ .
Since $y > z$ , we assign $y = 3$ and $z = 2$ .

$n = 2^{x} 3^{3} 5^{2}$

Substitute $y = 3$ and $z = 2$ back into $n$ .
$n = 2^{x} \cdot 3^{3} \cdot 5^{2}$
To find odd divisors, we must exclude any factors of $2$ .
We only consider the odd prime factors: $3^{3} \cdot 5^{2}$ .

$(a + 1) (b + 1)$

For a number $N = p^{a} q^{b}$ , the total number of divisors is $(a + 1) (b + 1)$ .
For our odd part: $3^{3} \cdot 5^{2}$
The exponent of $3$ is $a = 3$ .
The exponent of $5$ is $b = 2$ .

$(3 + 1) (2 + 1) = 12$

Number of odd divisors = $(3 + 1) \times (2 + 1)$
Number of odd divisors = $4 \times 3$
Number of odd divisors = $12$
The correct option is $12$ .

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The Sigma Insight: Combinations and Selection

Welcome, future engineers! Today, we embark on a journey into the heart of number theory, a field that might seem abstract but is the bedrock of modern cryptography and computational science.

We are looking at a number

n = 2^{x} 3^{y} 5^{z}

. Our mission is to find the number of odd divisors. This is not just about plugging in numbers; it is about understanding the structure of integers.

Analyzing the Constraints

We start with the constraints

y + z = 5

and

y^{- 1} + z^{- 1} = \frac{5}{6}

. The first step is to simplify the reciprocal equation.

By finding a common denominator, we transform

\frac{1}{y} + \frac{1}{z}

into

\frac{y + z}{y z}

. Since we know

y + z = 5

, this becomes:

\frac{5}{y z} = \frac{5}{6}

This is a moment of pure mathematical elegance! The numerators cancel out, leaving us with

y z = 6

. Now, we have the sum and the product of two numbers.

The Quadratic Bridge

This is the classic setup for a quadratic equation

t^{2} - (s u m) t + (p r o d u c t) = 0

. Substituting our values, we get:

t^{2} - 5 t + 6 = 0

Factoring this, we find the roots are

2

and

3

. Given the condition

y > z

, we must assign

y = 3

and

z = 2

The Odd Divisor Calculation

Now, we return to our original number

n = 2^{x} 3^{3} 5^{2}

. The question asks for the number of odd divisors.

An odd divisor cannot contain any factor of

2

. Therefore, we ignore the

2^{x}

term entirely and focus on the remaining part:

3^{3} 5^{2}

The number of divisors of a number

p^{a} q^{b}

is given by the formula

(a + 1) (b + 1)

. Applying this to our exponents

3

and

2

(3 + 1) (2 + 1) = 4 \times 3 = 12

The total number of odd divisors is 12.

Keep practicing, stay curious, and remember that every complex problem is just a collection of simple, beautiful steps waiting to be uncovered.

Common Traps & FAQs

Wait, why do we ignore the

2^{x}

term when counting odd divisors?

An odd number, by definition, cannot have any factor of

2

. If a divisor contains a factor of

2

, it becomes an even number. Therefore, to count only the odd divisors of

n = 2^{x} \cdot 3^{y} \cdot 5^{z}

, we must completely exclude the

2^{x}

term and focus only on the odd prime factors

3^{y}

and

5^{z}

Is there a faster way to find

y

and

z

without the quadratic equation?

While you might be able to guess the values

2

and

3

by inspection because they are small integers, the quadratic method is the most robust and systematic approach. In JEE Advanced, relying on intuition is risky; having a rigorous algebraic method ensures you never miss a solution or get confused when the numbers are larger or non-integers.

What if the question asked for the total number of divisors instead of just the odd ones?

That is a great question! If we needed the total number of divisors, we would include the

2^{x}

term. The formula would then be

(x + 1) (y + 1) (z + 1)

. Since the value of

x

is not provided, the question would have to give us more information, but the logic remains the same: add one to each exponent and multiply them together.

Animated Solution for Mathematics - Permutations and Combinations: A natural number has prime factorization given by $n = 2^{x} 3^{y} 5^{z}$ , where $y$ and $z$ are such that $y + z = 5$ and $y^{- 1} + z^{- 1} = \frac{5}{6}, y > z$ . Then the number of odd divisors of $n$ , including 1, is :

Select Answer:

Visualized Solution

$n = 2^{x} 3^{y} 5^{z}$

$y + z = 5$ and $\frac{1}{y} + \frac{1}{z} = \frac{5}{6}$

$\frac{y + z}{y z} = \frac{5}{6}$

$\frac{5}{y z} = \frac{5}{6}$

$y z = 6$

$t^{2} - 5 t + 6 = 0$

$y = 3, z = 2$

$n = 2^{x} 3^{3} 5^{2}$

$(a + 1) (b + 1)$

$(3 + 1) (2 + 1) = 12$

The Sigma Insight: Combinations and Selection

Analyzing the Constraints

The Quadratic Bridge

The Odd Divisor Calculation

Similar Questions

Let $S = {1, 2, 3, ....., 9}$ . For $k = 1, 2, ....., 5$ , let $N_{k}$ be the number of subsets of S, each containing five elements out of which exactly $k$ are odd. Then $N_{1} + N_{2} + N_{3} + N_{4} + N_{5} =$

Let $S = {p_{1}, p_{2} ...., p_{10}}$ be the set of first ten prime numbers. Let $A = S \cup P$ , where P is the set of all possible products of distinct elements of S. Then the number of all ordered pairs $(x, y)$ , $x \in S$ , $y \in A$ such that x divides y is

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For $n \geq 2$ , let $S_{n}$ denote the set of all subsets of ${1, 2, \dots, n}$ with no two consecutive numbers. For example ${1, 3, 5} \in S_{6}$ , but ${1, 2, 4} \in / S_{6}$ . Then $n (S_{5})$ is equal to ________

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Let $S = {1, 2, 3, ...., 100}$ . The number of non-empty subsets A of S such that the product of elements in A is even is :

Let $S = {1, 2, 3, 5, 7, 10, 11}$ . The number of non-empty subsets of $S$ that have the sum of all elements a multiple of 3, is ____.

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Let $S = {1, 2, 3, 4}$ . The total number of unordered pairs of disjoint subsets of $S$ is equal to

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=2x3y5z

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y+z=5 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y1​+z1​=65​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } yzy+z​=65​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } yz5​=65​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } yz=6

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t2−5t+6=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=3,z=2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=2x3352

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a+1)(b+1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (3+1)(2+1)=12

The Sigma Insight: Combinations and Selection

Analyzing the Constraints

The Quadratic Bridge

The Odd Divisor Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of triplets (x,y,z) where x,y,z are distinct non negative integers satisfying x+y+z=15, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of matrices of order 3×3, whose entries are either 0 or 1 and the sum of all the entries is a prime number, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={1,2,3,....,100}. The number of non-empty subsets A of S such that the product of elements in A is even is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={1,2,3,5,7,10,11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is ____.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={1,2,3,4,5,6,9}. Then the number of elements in the set T={A⊆S:A=ϕ and the sum of all the elements of A is not a multiple of 3} is .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={1,2,3,4}. The total number of unordered pairs of disjoint subsets of S is equal to

$n = 2^{x} 3^{y} 5^{z}$

$y + z = 5$ and $\frac{1}{y} + \frac{1}{z} = \frac{5}{6}$

$\frac{y + z}{y z} = \frac{5}{6}$

$\frac{5}{y z} = \frac{5}{6}$

$y z = 6$

$t^{2} - 5 t + 6 = 0$

$y = 3, z = 2$

$n = 2^{x} 3^{3} 5^{2}$

$(a + 1) (b + 1)$

$(3 + 1) (2 + 1) = 12$

The number of triplets $(x, y, z)$ where $x, y, z$ are distinct non negative integers satisfying $x + y + z = 15$ , is

The number of matrices of order $3 \times 3$ , whose entries are either 0 or 1 and the sum of all the entries is a prime number, is

Let $S = {1, 2, 3, ...., 100}$ . The number of non-empty subsets A of S such that the product of elements in A is even is :

Let $S = {1, 2, 3, 5, 7, 10, 11}$ . The number of non-empty subsets of $S$ that have the sum of all elements a multiple of 3, is ____.

Let $S = {1, 2, 3, 4, 5, 6, 9}$ . Then the number of elements in the set $T = {A \subseteq S : A \neq = ϕ and the sum of all the elements of A is not a multiple of 3}$ is .

Let $S = {1, 2, 3, 4}$ . The total number of unordered pairs of disjoint subsets of $S$ is equal to