Question

Shown below is a strip of paper which is folded multiple times. How many red pawns are placed on the same side of the paper as the blue pawn? 

Hint:

To solve paper folding problems, visualize each fold carefully to track how objects like pawns are shifted across different layers.

Answer 1

The problem requires determining how many red pawns are placed on the same side of the paper as the blue pawn in the given figure. 

Analysis 
The figure represents a strip of paper folded multiple times, creating a series of alternating sides for the red and blue pawns. To determine how many red pawns are on the same side as the blue pawn, we must carefully observe the placement of the pawns relative to the blue pawn. 

Steps to Solve 
The blue pawn is placed on one specific side of the folded paper. Every alternate pawn on the paper will be on the opposite side due to the folding pattern. 
Starting from the blue pawn, we trace along the strip of paper and identify the red pawns that are placed on the same side as the blue pawn.

Counting the Red Pawns 
Observing the placement in the figure, there are a total of 9 red pawns on the same side of the paper as the blue pawn. 
The remaining red pawns are on the opposite side.

Conclusion 
The number of red pawns on the same side of the paper as the blue pawn is 9.

Answer 2

The strip of paper zig-zags back and forth across the page, folding at each corner, with red pawns and one blue pawn printed at fixed points along its length. When such a strip is folded, every crease flips the strip over, so a pawn's face ends up on the same side as the blue pawn only if an even number of creases separates the two pawns along the strip.

Start at the blue pawn and walk along the strip in both directions, counting the fold lines (creases) crossed before reaching each red pawn.
Every time you cross one crease, the layer flips to the opposite face; cross a second crease, and it flips back to match the blue pawn's face again.
So a red pawn shares the blue pawn's side whenever an even number of creases lies between it and the blue pawn, and it lands on the opposite side whenever an odd number of creases lies between them.
Working through the full loop this way, an even number of creases separates the blue pawn from 9 of the red pawns, while the rest are separated by an odd number of creases and end up on the far side.

So the number of red pawns on the same side of the paper as the blue pawn is 9.