finding numbers

Question

adam's house number has three digits. if the leftmost digit of this number is removed, the resulting number is the house of ben. if the leftmost digit of ben's house number is removed, the resulting number is the house of paul. the sum of he three house numbers is 912. what is the hundreds' place digit of adam's house number?

Answer 1

Hi Tyler,

 

A good way to go about figuring this problem is denote Adam's house number by abc, where each of a, b, and c are numbers between 0 and 9. So, if Adam's house is abc, Ben's house is bc, and Paul's house is c. Then, we know that their sum is 912. So let's set this up:

 

 

      abc

        bc

+       c

_______

      912

 

OK. So, if we think about this problem like regular old addition, we have c+c+c would leave a remainder of 2 when divided by 10. In other words, c+c+c=3c would have to have a 2 in the ones digit. Numbers like this are 2, 12, 22, 32, 42, 52, 62, 72, 82, 92, etc. But since c is an integer (a whole number), 3c would also have to be a whole number. So in our list, we need to look at numbers divisible by 3. This leaves 12, 42, and 72. If 3c=12, then c=4. If 3c=42, then c=14, but c cannot be 14 because c should be between 0 and 9. Hence, c=4. Our problem now becomes:

 

        ab4

          b4

+         4

________

         912

 

OK, we know 4+4+4 is 12, so adding in the usual way makes us leave the 2 below the equal sign, and carry the one above the b, like this:

 

        a¹b4

           b4

+           4

________

         912

 

Alright. Now, we need to have b+b+1 have a one when divided by 10, so b+b+1=2b+1 could equal 11, 21,31,41,51,61,71,81,91,etc... But if 2b+1=21, then b would have to be 10. But b must be between 0 and 9, so we choose 2b+1=11, and so b=5. Almost done. Now our problem is

 

         a54

           54

+           4

_________

         912

 

OK, but when we add 4+4+4, we get 12, so we carry the one to the fives. But 5+5+1=11, so we again need to carry the 1 over to a. Then, a+1=9, and so a=8.

 

Hence, our number for Adam's house is 854, Ben's is 54, and Paul's is 4.

 

Hope this helps.

Answer 2

Adam's house number is abc

Ben's house number is bc

Paul's house number is c

abc=100a+10b+c

bc=10b+c

c=c

The sum of the three numbers is 912.

100a+10b+c+10b+c+c=912

100a+20b+3c=912

a cannot be 9 because 912-900=12 and therefore there is no digit for b

if a is 8, then 912-800=112

20b+3c=112

b can be 5 because 20*5=100 and 112-100=12

3c=12 and c=4

the numbers are 854, 54, and 4

854+54+4=912