Let the sum of two positive integers be 24 . If the probability, that their product is not less than

Q: Why is the sample space 23 and not 24?

Great question! Since $x$ and $y$ are positive integers, $x$ cannot be 0. If $x=24$, then $y=0$, which is not allowed. Thus, $x$ must be at least 1 and at most 23. This gives us exactly 23 possible pairs.

Q: Why does the maximum product occur at $x=12$?

Think of the product $P(x) = x(24-x)$ as a parabola. The roots are 0 and 24. By symmetry, the vertex is exactly at the midpoint, which is 12. Alternatively, you can differentiate $P(x)$ with respect to $x$ and set it to zero.

Q: How do I count the integers in the range $6 \le x \le 18$?

When you have an inclusive range $[a, b]$, the number of integers is $b - a + 1$. Here, $18 - 6 + 1 = 13$. It is a common mistake to just subtract the endpoints, but remember to add one because both 6 and 18 are included!

Question

Let the sum of two positive integers be 24 . If the probability, that their product is not less than $\frac{3}{4}$ times their greatest possible product, is $\frac{m}{n}$, where $gcd(m, n) = 1$, then $n - m$ equals

Options:

(A) 10

(B) 9

(C) 11

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Constraints

We are given two positive integers, $x$ and $y$, such that their sum is fixed at $x + y = 24$. Since $x$ and $y$ must be positive integers, $x$ can take any integer value from $1$ to $23$.

The total number of possible outcomes in our sample space is therefore $23$.

### The Product Function

We define the product $P = xy$. By substituting $y = 24 - x$, we express the product as a function of $x$:

$$P(x) = x(24 - x) = 24x - x^2$$

This is a downward-facing parabola. The maximum value of this product occurs at the vertex, which is located at $x = 12$.

Substituting $x = 12$ into the function, we find the maximum product:

$$P_{max} = 12(24 - 12) = 144$$

### Defining the Favorable Condition

The problem asks for the probability that the product is "not less than" $\frac{3}{4}$ of this maximum. This is equivalent to the condition $P(x) \ge \frac{3}{4} 	imes 144$.

Calculating the threshold value:

$$P(x) \ge 108$$

Substituting our expression for $P(x)$, we solve the inequality:

$$24x - x^2 \ge 108$$

### Solving the Inequality

Rearranging the terms to one side, we obtain:

$$x^2 - 24x + 108 \le 0$$

To factorize this quadratic, we look for two numbers that multiply to $108$ and add to $-24$. These numbers are $-6$ and $-18$.

The inequality becomes:

$$(x - 6)(x - 18) \le 0$$

For an upward-opening parabola, the expression is less than or equal to zero between the roots. Thus, the valid range for $x$ is $6 \le x \le 18$.

### Final Calculation

The number of favorable outcomes is the count of integers in the set $\{6, 7, 8, \dots, 18\}$. Using the formula for the number of terms in an inclusive range:

$$	ext{Count} = 18 - 6 + 1 = 13$$

The probability is the ratio of favorable outcomes to the total sample space:

$$	ext{Probability} = \frac{13}{23}$$

Given that $m = 13$ and $n = 23$ are coprime, the final result is:

$$n - m = 23 - 13 = \mathbf{10}$$

Animated Solution for Mathematics - Probability: Let the sum of two positive integers be 24 . If the probability, that their product is not less than $\frac{3}{4}$ times their greatest possible product, is $\frac{m}{n}$ , where $g c d (m, n) = 1$ , then $n - m$ equals

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Visualized Solution

$x + y = 24$

$Sample Space$

$P (x) = x (24 - x)$

$P_{ma x} = 144$

$P (x) \geq \frac{3}{4} P_{ma x}$

$P (x) \geq 108$

$x^{2} - 24 x + 108 \leq 0$

$(x - 6) (x - 18) \leq 0$

$6 \leq x \leq 18$

$n (E) = 13$

$P (E) = \frac{13}{23}$

$n - m = 10$

The Sigma Insight: Classical Definition of Probability

Analyzing the Constraints

The Product Function

Defining the Favorable Condition

Solving the Inequality

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+y=24

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x)=x(24−x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pmax​=144

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x)≥43​Pmax​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x)≥108

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2−24x+108≤0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x−6)(x−18)≤0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6≤x≤18

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(E)=13

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E)=2313​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n−m=10