Peter has invented a game with paper cups. He lines up 121 cups face down in a straight line from left to right and consecutively labels them from 1 to 121. He then walks from left to right down the line of cups, flipping all of the cups over. He returns to the left end of the line, then makes a second pass from left to right, this time flipping cups 2,4,6,8... On the third pass, he flips cups 3,6,9,12 He continues like this: On the ith pass, he flips over cups i, 2i, 3i, 4i, (By "flip," we mean changing the cup from face down to face up or vice versa.) After 121 passes, how many cups are face up?

After 121 passes, there will be 11 cups facing up

Step-by-step explanation:

Given that:

Peter initially  lines up 121 cups facing down in a straight line from left to right and consecutively labels them from 1 to 121.

We can have an inequality ; i.e 1 ≤ n ≤  121; if n represents the divisor including n itself for which n  = odd number. Thus at the end of this claim, the cup will be facing up.

On the ith pass, he flips over cups i, 2i, 3i, 4i,.... (By "flip," we mean changing the cup from face down to face up or vice versa.)

For each divisor on the ith pass of n;

 since we are dealing with possibility of having an odds number:

Thus; and  where ; n = perfect square.

Thus ; we will realize that between 1 to 121 ; there exist 11 perfect squares. Therefore; as a result of that ; 11 cups will definitely be facing up after 121 passes

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step-by-step explanation:

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