Math Factor Hopper for Windows: Game Display

This is a screen shot showing the game in play in Level 2. Your goal in the game is to make the sum of all the numbers on the board as small as possible, and you do this by "factor hopping" one frog over another. The hopping frog hops over a factor of itself, the factor is removed, and the result is multiplied into the destination frog. You want to get the SMALLEST sum in the FEWEST hops! Your hops are tallied to the right of the board (hopping frogs keep count of your "hop count"). You can replay a game level to try to better your score, or you can just move up a level as you complete each game board.

When you go up a level the game board gets bigger and the factors get more complex.

Examine the above screen shot; learn these four rules and read on for a few example turns based on the above game board.

Rules are simple:

• Green frogs can hop.
• Blue frogs can't hop.
• Frogs can only hop over numbers which are their factor.
• Frogs can hop up, down, left, right or diagonally

If a Green frog lands on a Green frog, it stays green. If a Green frog lands on a Blue frog, it becomes blue. Whenever a frog becomes a 1 frog, it becomes a blue frog. Let's just play a few turns... YOU'LL LEARN THE RULES AS YOU PLAY! Some folks play and ignore the rules!

Let's hop the active frog (the frog on the red lily pad) up over the frog above it (the middle yellow arrow, pointing up). This gives us the following board:

The 4 frog hopped over a 2 frog onto a 1 frog. The result was that the 2 factor was sucked out of the hopping frog as it hopped over the 2 frog, leaving 2 in the hopping frog (4 / 2 = 2) and then when the frog landed on the 1 frog it converted it to a 2 frog (2 x 1 = 2).

Next, let's hop the big 16 frog over the 2 frog we just made (the top arrow):

This gives us the following board:

Remember how the game works? The hopping frog hops over a factor of itself, the factor is removed, and the result is multiplied into the destination frog. We hopped the 16 frog over the 2 frog, leaving 8 in the frog (16 / 2 = 8). We landed on a 1 frog, so the 1 frog became an 8 frog (1 x 8 = 8).

Did we do something dumb? We now have an 8 frog which is blue! We can't hop with blue frogs, so are we going to be stuck with an 8 frog at the end of this level? Nope. We can hop the 4 frog on the bottom row up diagonally onto the 2 from on the middle row. This gives us 8 on a green frog (where the 2 frog was - he got converted to an 8 frog when the 4 frog landed on him). We can then hop the 8 frog (Green frog) up over the 8 blue frog, and we got rid of the two 8 frogs!

Every board has a different solution, and there are an unlimited number of plays!

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