6
$\begingroup$

The number of all 3-digit numbers abc (in base 10) for which $\ abc+ab+bc+ac+a+b+c = 29$ is
(A) 6
(B) 10
(C) 14
(D) 18

My working:
$\ ab (c + 1) +b (c + 1) + a(c + 1) + c + 1 = 30$
$\ (a + 1) (b +1) (c+1) = 30 $
$\ 9>a>0$
$\ 0\le b,c<9$

The problem I am facing:
I don't know how to figure out the no. of solutions for which this is true.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ The title is a bit confusing as it contains "abc" in two different meanings. $\endgroup$ – JiK Aug 8 '16 at 18:08
  • $\begingroup$ @JiK. The title and the Q are also misleading as none of the solutions include what are usually called "3-digit numbers in base ten". $\endgroup$ – DanielWainfleet Aug 8 '16 at 19:19
  • 1
    $\begingroup$ @user254665, both title and question are asking how many 3-digit numbers have a specified, rather constraining, property, so it makes sense that the answer is less than $100$ (as the posted solutions show to be the case). $\endgroup$ – Barry Cipra Aug 9 '16 at 0:42
17
$\begingroup$

$$(a+1)(b+1)(c+1)=30$$

Total number of ways to satisfy above equation:

  1. $30 = (2,3,5)$ gives $6$ values for $a+1,b+1,c+1$ $(i.e., a=1, b=2, c=4)$
  2. $30 = ( 1 , 6 ,5)$ gives $6$ values for $a+1,b+1,c+1$ $(i.e., a=0, b=5, c=4)$
  3. $30 = (1,3,10)$ gives $6$ values for $a+1,b+1,c+1$ $(i.e., a=0, b=2, c=9)$

In the first case we have $6$ values for $a,b,c$ as $(1,2,4)$ can take up any digit hence they can be arranged in $3!$ ways.

In the second case we have $4$ values for $a,b,c$ as $(0,5,4)$. $0$ can not be as first digit as. Then $abc$ would be a $2$ digit number ( contrary to given condition ). Hence we have $ 2.2.1$ number of ways .

Similarly in the third case we have $4$ values $a,b,c$ as $(0,2,9)$. $0$ can not be as first digit as. Then $abc$ would be a $2$ digit number ( contrary to given condition ). Hence we have $ 2.2.1$ number of ways .

Therefore, total number of ways:

$$3!+2\times 2\times1+2\times 2\times1=6+4+4=14 \space \text{ways}$$

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ Perhaps you could combine this with Jennifer's answer to justify why these are the only possibilities. $\endgroup$ – Morgan Rogers Aug 8 '16 at 15:23
3
$\begingroup$

The prime fctorisation of $30$ is $5\times3\times2$. The only ordered 3-uples for $a+1,b+1$ and $c+1$ where $a+1 \in [2,10]$ and $b+1,c+1 \in [1,10]$ are : $$(2,3,5),(2,5,3),(3,1,10),(3,2,5),(3,5,2),(3,10,1),(5,1,6),(5,2,3),(5,3,2),(5,6,1),(6,1,5),(6,5,1),(10,1,3),(10,3,1)$$

So the solutions are : $$124,142,209,214,241,290,405,412,421,450,504,540,902,920$$

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ Second last number should be 902 instead of 903. $\endgroup$ – Tejas Kale Aug 8 '16 at 18:34
2
$\begingroup$

A quick bruteforce approach to figuring out the possible number of solutions:

Javascript code, paste in browser console for a quick test.

for(i=100;i<=999;i++){
    a=i.toString();
    //Take the individual characters(the digits) of the string, increment by 1 and multiply.
    if(++a[0]*++a[1]*++a[2] === 30){
        console.log(i);
    }
}

This gives the following 14 numbers:
124 142 209 214 241 290 405 412 421 450 504 540 902 920

|cite|improve this answer
$\endgroup$
1
$\begingroup$

Another brute-force approach, using Python:

[100*a+10*b+c for a in range(1,10) for b in range(10) for c in range(10)\ if a*b*c+a*b+b*c+a*c+a+b+c == 29]

yields[124, 142, 209, 214, 241, 290, 405, 412, 421, 450, 504, 540, 902, 920]

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.