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: $6%20%2B%207%20%2B%208%20%2B%209%20%2B%2010%20%2B%209%20%2B%208%20%2B%207%20%3D%2064$

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Natural Number

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs