AS SIMPLE AS DYING]]>

This is what I did to solve this.

Suppose the number is abc where a is the number of hundreds, b the number of tens, and c the remaining units. All three must be singkle digit numbers ie from the set {0,1,2,3,4,5,6,7,8,9}

To express this correctly using algebra you have to find a way to show that the 'a' is worth 'a' hundreds and the 'b' is worth 'b' tens. So you write the number as 100a + 10b + c. Doing this makes sure that the correct place value is applied to the hundreds and the tens.

We also have to add on the actual digits so the total becomes 100a + 10b + c + a + b + c = 101a + 11b + 2c = 379

Now the only way you could get to 3 hundred and something is if a = 3. Less than 3 and the total would be under 300. More than that and the total would be over 400. So a= 3 is forced upon us.

So subtract 3a = 303 and the equation becones 11b + 2c = 379 - 303 = 76

Now you can apply a similar logic to determine 'b'. If it were b = 7 the equation would be 77 + 2c = 76. That cannot be right, as c would have to be negative.

And if it were only b = 5 the equation would be 55 + 2c = 76 and yet 'c' is 9 or less so that doesn't work either.

So b is forced to be ? I'll leave you to finish this off from here.

Bob

]]>I deleted my answer, which was just the number.

I posted it very prematurely without giving any help at all, despite this being the Help Me! forum.

Sorry about that!