Are we supposed to derive this without trial and error? I know what the answer is, but I don't know how to prove it algebraically. Ok here goes.
C+2 = B-2 = A.
B-4 = C.
B+1 = 2A.
4-1 = 3, but does that logically prove that A=3 and therefore that C=1 and B=5? Or do I just think it does because I tried a few values and worked out which of them was right?
The difference between the number of sweets A and B have is two. If B had one more sweet, he would have twice as many as A. Therefore, A + 3 = 2A. which means A has three sweets. B must then have five, as he has one less than twice what A has. If C had two more sweets he would have the same as A, so he must have one.
Oh ek! Give me a few years to get my head around that one...
They put logic puzzles in the classrooms and change them each week. I can do primary 4 most days which is about age 8/9 , further up the school it takes me an age, other times I get them as soon as I look at them. Maybe it's things I've seen before but don't realise.
This type of puzzle has me completely baffled. I'm better at word ones.
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