ok people, i have a question,
can 418 be spilt into groups of three where an integer is less than 180?
all of the integers have to be different and they must be in the format a+b+c= 480.
if there are any integers, please list them and explain why these work.
if not, prove that there can be no such integers of this kind.
Thanks!
(idk if this is a duplicate question but oh well if it is)
I wrote a short code and listed all 3-digit combinations and they came out to =7,260 !!. Here is the code:
a=1; b=1;d=1;p=0;c=a + b + d; if(c==418, goto6, goto8);p=p+1;printp," - ",c, a, b, d; a++;if(a<180, goto4, 0);a=1;b++;if(b<180, goto4, 0);a=1;b=2;d++;if(d<180, goto4,0)
P.S. I hope that he/she doesn't want them ALL listed!.
+9519 418 or 480?
And did you mean that whether 3 integers (a, b, and c) can be found such that \(a< b< c<180\) and \(a + b + c = 418\text{(or 480)}\)?
Yes, 418 can be split into a group of three numbers.
418 can be split into 138 + 139 + 141.
418/3 = 139.333333333 and that keeps gong on forever. So the numbers must be around that point somewhere.
138+139+140= 417. We are one number off from 418. So, logically, if we added 1 to one of the numbers, we would get 418.
Since you said all the numbers must be different, we can't add the one to 138 or 139, since then two numbers would be the same.
But, we could add the one two 140, since there are no other numbers in that equation that equal to "141."
So, now our equation is 138+139+141= 418.
Hope this helps!
I wrote a short code and listed all 3-digit combinations and they came out to =7,260 !!. Here is the code:
a=1; b=1;d=1;p=0;c=a + b + d; if(c==418, goto6, goto8);p=p+1;printp," - ",c, a, b, d; a++;if(a<180, goto4, 0);a=1;b++;if(b<180, goto4, 0);a=1;b=2;d++;if(d<180, goto4,0)
P.S. I hope that he/she doesn't want them ALL listed!.
