Is there an easier way to add up the 3-digit numbers?

[davidtrinh]

Hi everyone. I am a new member. I do math as a hobby. I am not sure how to do this problem.

(ex) Determine the sum of all the 3-digit numbers that can be made by choosing 3 different numbers from 1, 2, 3, 4, 5, 6, 7.

This is how I do it.

With the first three numbers 1, 2 , 3, I get six 3-digit ones: 123 132 213 231 312 321.

Adding them gives 1332.

With the first four numbers 1, 2, 3, 4, I get twenty four 3-digit ones: 123 124 134 132 142 143 213 214 234 231 241 243 312 314 324 321 341 342 412 413 423 421 431 432.

Adding them gives 6660.

My way is too slow. Is there a faster way of solving the problem? Please help me. Thanks.

 

[mmm4444bot]

Hi everyone. I am a new member. I do math as a hobby. I am not sure how to do this problem.

(ex) Determine the sum of all the 3-digit numbers that can be made by choosing 3 different numbers from 1, 2, 3, 4, 5, 6, 7.

...

My way is too slow. Is there a faster way of solving the problem?

Welcome!

I'm not sure whether writing a program would be faster, but do you have programming skills? This exercise could be coded in BASIC. :cool:

 

[pka]

Hi everyone. I am a new member. I do math as a hobby. I am not sure how to do this problem. (ex) Determine the sum of all the 3-digit numbers that can be made by choosing 3 different numbers from 1, 2, 3, 4, 5, 6, 7.

This is how I do it.

In the list of the three distinct digits numbers each of \(\displaystyle 1,2,3,4,5,6,7\) appears thirty times in each column.
The sum of \(\displaystyle 1+2+3+4+5+6+7=28\)
Thus the sum of each column \(\displaystyle 30(28)\).
Therefore, the sum of all numbers \(\displaystyle 30(28)(10^2)+30(28)(10^1)+30(28)(10^0)=~?\).

 

[davidtrinh]

Hi pka. I am confused about your strategy. Why do you add up the seven numbers? How do you know the 3 distinct digit numbers appear 30 times in each column? Could you please explain them? Thank you.

 

[davidtrinh]

I know pka's strategy is correct. It is an excellent solution. I am not challenging him. But, I don't understand why you have to add up the seven numbers? How do you know the distinct 3-digit numbers appear 30 times in each column?

 

[pka]

I know pka's strategy is correct. It is an excellent solution. I am not challenging him. But, I don't understand why you have to add up the seven numbers? How do you know the distinct 3-digit numbers appear 30 times in each column?

Using only the digits six \(\displaystyle 2,3,4,5,6,7\) form two digit numbers \(\displaystyle XY\) where \(\displaystyle X\ne Y\). There are six choices for \(\displaystyle X\) and five choices for \(\displaystyle Y\). That is a total thirty, two digit numbers having distinct digits. Put a \(\displaystyle 1\) in the hundredth position and now we have thirty, three digit numbers having distinct digits. Doing that for each the others gives a total of \(\displaystyle 210\) three digit numbers having distinct digits.

 

[Steven G]

Hi everyone. I am a new member. I do math as a hobby. I am not sure how to do this problem.

(ex) Determine the sum of all the 3-digit numbers that can be made by choosing 3 different numbers from 1, 2, 3, 4, 5, 6, 7.

This is how I do it.

With the first three numbers 1, 2 , 3, I get six 3-digit ones: 123 132 213 231 312 321.

Adding them gives 1332.

With the first four numbers 1, 2, 3, 4, I get twenty four 3-digit ones: 123 124 134 132 142 143 213 214 234 231 241 243 312 314 324 321 341 342 412 413 423 421 431 432.

Adding them gives 6660.

My way is too slow. Is there a faster way of solving the problem? Please help me. Thanks.

Well yeah there is an easier way. In 1, 2, 3 , 4 you obviously have 1, 2 and 3. The 1332 cases you got from just using 1, 2, 3 are counted again in your 6660. Just stop counting duplicates and it will be done faster. I suspect that there are much faster ways.

 

[davidtrinh]

Thanks to you all. Many thanks to pka.