An algebra problem by Sugam Bhandari
Algebra
Level
pending
I plan on investing shares each worth $23, $27, $45 and $21. If I buy at least one of each share, and spend exactly $215, how many shares did I buy altogether?
The answer is 7.
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1 solution
Let a, b, c and d be thr number of $21, $45, $27 and $23 respectively. Therefore 21a + 45b + 27c + 23d = 215. Therefore d = \frac {215 - 3 (7a+15b+9c)}{23}. Let (7a+15b+9c)=v, we get d= \frac {215-3v}{23}. Ie (215-3v) =23,46,69,92.... through a bit of trial and error, we get 215-3v=23 or 92. When its 92, v=41, ie 7a+15b+9c=41 and making a the subject a = frac{41-3 (5b+3c)}{7}. Again let (5b+3c)=w therefore a= frac {41-3w}{7} since w cannot equal 2 as I bought each share, the next possibility is when w= 9. However there is no Integer solution for 5a+3b=14 therefore v= 41 doesnt work. Therefore when v= 64, 7a = 64-3y where y = 5b+3c. For the reasons above when y=5 and 12 integer solutions dont exist. However when y=19, b=2, c=3, d=1 and a=1, ie a+b+c+d= 2+3+1+1= 7. You can also check that these values satisfy.