Natural Number

Let the natural number be represented as $x%20%3D%20%5Coverline%7Bxyz%7D$ , where $x%2C%20y%2C%20z$ are the digits of the number. We are given that: $x%20%2B%20y%20%2B%20z%20%3D%2015$ We need to find all valid combinations of $x%2C%20y%2C%20z$ where the sum equals 15 and $x$ is the hundreds digit (i.e., $2%20%5Cleq%20x%20%5Cleq%209$ ).
Step 1: Case for $x%20%3D%202$ : Here, $y%20%2B%20z%20%3D%2013$ , and the possible pairs for $y$ and $z$ are: $(4%2C%209)%2C%20(5%2C%208)%2C%20(6%2C%207)%2C%20(7%2C%206)%2C%20(8%2C%205)%2C%20(9%2C%204)$ Thus, there are 6 possibilities.
Step 2: Case for $x%20%3D%203$ : Here, $y%20%2B%20z%20%3D%2012$ , and the possible pairs for $y$ and $z$ are: $(3%2C%209)%2C%20(4%2C%208)%2C%20(5%2C%207)%2C%20(6%2C%206)%2C%20(7%2C%205)%2C%20(8%2C%204)%2C%20(9%2C%203)$ Thus, there are 7 possibilities.
Step 3: Case for $x%20%3D%204$ : Here, $y%20%2B%20z%20%3D%2011$ , and the possible pairs for $y$ and $z$ are: $(2%2C%209)%2C%20(3%2C%208)%2C%20(4%2C%207)%2C%20(5%2C%206)%2C%20(6%2C%205)%2C%20(7%2C%204)%2C%20(8%2C%203)%2C%20(9%2C%202)$ Thus, there are 8 possibilities.
Step 4: Case for $x%20%3D%205$ : Here, $y%20%2B%20z%20%3D%2010$ , and the possible pairs for $y$ and $z$ are: $(1%2C%209)%2C%20(2%2C%208)%2C%20(3%2C%207)%2C%20(4%2C%206)%2C%20(5%2C%205)%2C%20(6%2C%204)%2C%20(7%2C%203)%2C%20(8%2C%202)%2C%20(9%2C%201)$ Thus, there are 9 possibilities.
Step 5: Case for $x%20%3D%206$ : Here, $y%20%2B%20z%20%3D%209$ , and the possible pairs for $y$ and $z$ are: $(0%2C%209)%2C%20(1%2C%208)%2C%20(2%2C%207)%2C%20(3%2C%206)%2C%20(4%2C%205)%2C%20(5%2C%204)%2C%20(6%2C%203)%2C%20(7%2C%202)%2C%20(8%2C%201)%2C%20(9%2C%200)$ Thus, there are 10 possibilities.
Step 6: Case for $x%20%3D%207$ : Here, $y%20%2B%20z%20%3D%208$ , and the possible pairs for $y$ and $z$ are: $(0%2C%208)%2C%20(1%2C%207)%2C%20(2%2C%206)%2C%20(3%2C%205)%2C%20(4%2C%204)%2C%20(5%2C%203)%2C%20(6%2C%202)%2C%20(7%2C%201)%2C%20(8%2C%200)$ Thus, there are 9 possibilities.
Step 7: Case for $x%20%3D%208$ : Here, $y%20%2B%20z%20%3D%207$ , and the possible pairs for $y$ and $z$ are: $(0%2C%207)%2C%20(1%2C%206)%2C%20(2%2C%205)%2C%20(3%2C%204)%2C%20(4%2C%203)%2C%20(5%2C%202)%2C%20(6%2C%201)%2C%20(7%2C%200)$ Thus, there are 8 possibilities.
Step 8: Case for $x%20%3D%209$ : Here, $y%20%2B%20z%20%3D%206$ , and the possible pairs for $y$ and $z$ are: $(0%2C%206)%2C%20(1%2C%205)%2C%20(2%2C%204)%2C%20(3%2C%203)%2C%20(4%2C%202)%2C%20(5%2C%201)%2C%20(6%2C%200)$ Thus, there are 7 possibilities. Now, the total number of possible values is: $6%20%2B%207%20%2B%208%20%2B%209%20%2B%2010%20%2B%209%20%2B%208%20%2B%207%20%3D%2064$

About Natural Number - JEE-MAIN

Natural Number is a vital chapter for JEE-MAIN aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.

By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.

Frequently Asked Questions

Why focus on Natural Number PYQs?

Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.

How to best use this analysis?

Review the topic breakdown to see which sub-topics within Natural Number carry the most weight. Then, tackle the questions iteratively to solidify your understanding.

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

About Natural Number - JEE-MAIN

Frequently Asked Questions

Why focus on Natural Number PYQs?

How to best use this analysis?

Natural Number

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

About Natural Number - JEE-MAIN

Frequently Asked Questions

Why focus on Natural Number PYQs?

How to best use this analysis?

Chapter Questions
2 MCQs