An algebra problem by Sagar Saluja
Algebra
Level
pending
If a,b and c are integers and a + b + c +ab + bc +ca +abc = 998 Then what will be the value of X if X= a+b+c /19 where X belongs to Integers.
6
4
1
0
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1 solution
We have a + b + c +ab +bc + ac +abc =998
so it can be written as
a + ac +b + bc +ab +abc +c +1 -1 =998
a(1 + c) + b(1+c) +ab (1+c) + 1(1+c) -1 =998
(a + b +ab +1)(1+c) =999
(a + b +ab +1)(1+c) = 1000 -1
(a + b +ab +1)(1+c) = (10 - 1)(100 +1+10)
(a + b +ab +1)(1+c) = (9) (111)
Case 1:
if (a + b +ab +1)=111 c+1 =9
a(1+b) +1=111 c=8
(a+1)(b+1)=37 3
a=36 and b=2 or
a=2 and b=36
so a+ b +c =44
but 44/19 is not an integer.
So we proceed to case 2.
Case 2:
if (a + b +ab +1)=9 and c+1 =111
a(1+b) +1=9 c =110
(a+1)(b+1)=3 3
a=2
b=2
a+b+c =2+2+110 =114
and a +b +c /19 =6 which is an integer.