Digit addition problem

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Digit addition problem

by contactsharma » Mon Dec 29, 2008 11:16 pm
1234
1243
1324
.......
.......
+4321

Q. The addition problem above shows four of the 24 different integers that can be formed usinng each of the digitd 1,2,3,4 exactly once in each integer. What is the sum of these 24 integers?

Its an OG 10 problem, so I know the answer but I have failed to understansd the explanation given. Please explain the process of arriving at the answer in detail.
Source: — Problem Solving |

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by brb588 » Mon Dec 29, 2008 11:53 pm
The problem helps you by letting your know their is only 24 possibilities. Start with the thousands digit: If you figure it out by hand, you can see that only six numbers begin with 1. You can infer that six begin with 2, six with 3, and six with 4.

1000*6 + 2000*6 + 3000*6 + 4000*6 =

10000*6 =

60000

Now to the hundreds. Six of the twenty-four numbers have 1 in the hundreds place, and so on...

100*6 + 200*6 + 300*6 + 400*6 =

1000*6 =

6000

It should be obvious that we have have a pattern. Tens digit should give me 600, ones digit gives me 60. Now add them up.

60000 + 6000 + 600 + 60 = 66660

If I'm misguided, please someone let me know.

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digit addition problem

by contactsharma » Tue Dec 30, 2008 9:12 am
thanks..that makes it so much simpler :)

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by rajataga » Tue Dec 30, 2008 11:04 am
i used the same concept, but solved it in a slightly different way.....

since there are 24 terms existing in all, and in each term, every number is unique, so each of the number exists in the same 6 times....

now add 4 + 3 + 2 + 1 = 10, since all the numbers appear 6 times, multiple it by 6......you get 60.

now, when adding 2-3 terms we list them down one below the other and add them, rite, think of this in a similar way...

the units digit will add up to 60, 6 gets carried over above the tens digit. tens will also add up to 60 + 6 = 66

hence, you get the answer, 66660