- What is the total number of 3-digit numbers, whose sum of digits is 10?

Question

Vedantu Content Team · Accepted Answer

- Hint: Let the three digits be $$x,y,z$$. Clearly, $$1 \le x \le 9,0 \le y \le 9,0 \le z \le 9$$. Then use the rule of expansion of a 3-digit number and make an equation following the condition “sum of the three digits is 10”. After that put the different values of $$x,y,z$$ and find out the required answer.Formula used: The expansion of a 3 digit number is $$100x + 10y + z$$, where $$x,y,z$$ are the digits of the number at hundred’s place, ten’s place and unit’s place respectively.Complete step-by-step solution: Let $$x,y,z$$ are the digits of a 3 digit number at hundred’s place, ten’s place and unit’s place respectively.Then the number in the form of expansion is $$100x + 10y + z$$.It is given that the sum of the digits is $$10$$.So, $$x + y + z = 10......\left( i \right)$$, where $$1 \le x \le 9,0 \le y \le 9,0 \le z \le 9$$If $$x = 1$$, then $$y + z = 9 \Rightarrow z = 9 - y$$Putting $$y = 0$$, we get $$z = 9$$Putting $$y = 1$$, we get $$z = 8$$Putting $$y = 2$$, we get $$z = 7$$Putting $$y = 3$$, we get $$z = 6$$Putting $$y = 4$$, we get $$z = 5$$Putting $$y = 5$$, we get $$z = 4$$Putting $$y = 6$$, we get $$z = 3$$Putting $$y = 7$$, we get $$z = 2$$Putting $$y = 8$$, we get $$z = 1$$Putting $$y = 9$$, we get $$z = 0$$So, the numbers are $$109,118,127,136,145,154,163,172,181,190$$$$\therefore $$ From $$100$$ to $$199$$, there are total $$10$$ such numbers.Similarly, putting $$x = 2$$ in equation $$(i)$$, we get $$y + z = 8$$Putting $$y = 0,1,2,3,4,5,6,7,8$$ respectively, we get $$z = 8,7,6,5,4,3,2,1,0$$We can’t put $$y = 9$$ because putting $$y = 9$$, we get $$z =  - 1$$, which is not possible.So, the numbers are $$208,217,226,235,244,253,262,271,280$$$$\therefore $$ From $$200$$ to $$299$$, there are total $$9$$ such numbers.Similarly, putting $$x = 3$$, we get the numbers $$307,316,325,334,343,352,361,370$$$$\therefore $$From $$300$$ to $$399$$, there are total $$8$$ such numbers.Similarly, putting $$x = 4$$, we get the numbers $$406,415,424,433,442,451,460$$$$\therefore $$From $$400$$ to $$499$$, there are total $$7$$ such numbers.Similarly, putting $$x = 5$$, we get the numbers $$505,514,523,532,541,550$$$$\therefore $$From $$500$$ to $$599$$, there are total $$6$$ such numbers.Similarly, putting $$x = 6$$, we get the numbers $$604,613,622,631,640$$$$\therefore $$From $$600$$ to $$699$$, there are total $$5$$ such numbers.Similarly, putting $$x = 7$$, we get the numbers $$703,712,721,730$$$$\therefore $$From $$700$$ to $$799$$, there are total $$4$$ such numbers.Similarly, putting $$x = 8$$, we get the numbers $$802,811,820$$$$\therefore $$From $$800$$ to $$899$$, there are total $$3$$ such numbers.Similarly, putting $$x = 9$$, we get the numbers $$901,910$$$$\therefore $$From $$900$$ to $$999$$, there are total $$2$$ such numbers.Hence, we get total $$10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 = 54$$ such numbers.Note: Many student mistake while putting the values of $$x,y,z$$ and using the condition $$1 \le x \le 9,0 \le y \le 9,0 \le z \le 9$$ properly. Since, the digits must be of 3 digits, so the digit of hundred’s place should not be equal to zero.