Guessing, if done properly, can be used to construct an
equation that will take you directly to the answer.
The key is to defer evaluations during the checking phase,
as illustrated below:
Note that being able to jump to the general equation depends on
our being able to spot where our guess showed up in the specific
equation. If we had evaluated, e.g., 60/5 and replaced it by 12,
this becomes much harder.
Q: Bees are let out of a hive. One-fifth of the bees fly to the rosebush,
one-third fly to the apple tree, and three times the difference
fly to the acorn tree. One bee is left flying around. How many bees
are there altogether?
A: The number of bees will have to be divisible by both 3 and 5.
30 is divisible by both 3 and 5, so could that be the answer?
30/5 = 6 bees fly to the rose bush.
30/3 = 10 bees fly to the apple tree
3*(10 - 6) = 12 bees fly to the acorn tree
1 just flies around
Total = 6 + 10 + 12 + 1 = 29 bees
which is close, but not quite right, since we need to end up with 30.
So we didn't get the right answer. However, we got something almost
Note that if we write our sum this way,
30/5 + 30/3 + 3(30/3 - 30/5) + 1 = ?
we can just substitute other values for 30 to check them. For
example, 60 is also divisible by both 3 and 5, so we can test
60/5 + 60/3 + 3(60/5 - 60/3) + 1 =
12 + 20 + 3(20 - 12) + 1 =
32 + 24 + 1 = 57
So that doesn't work either, but we can use this expression to check
answers very quickly. So at this point, we might start examining the
possibilities systematically, starting with 15, 30, 45, 60, and so on,
until we come across the answer.
But we can be a little more clever than that. Note that we'd like to
end up with our guess on the right side of the equals sign, e.g.,
30/5 + 30/3 + 3(30/3 - 30/5) + 1 =? 30
Suppose we replace the guess, wherever it appears, with an unknown
quantity, say, 'U'. Then we end up with this:
U/5 + U/3 + 3(U/3 - U/5) + 1 = U
Now, instead of using arithmetic to check guesses, we can use algebra
to find the correct 'guess' directly.