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Six identical cards are placed on a table. Each card has the number ‘1’ marked on one side and the number ‘2’ marked on its other side. All the six cards are placed in such a manner that the number ‘1’ is on the upper side. In one try, exactly four (neither more nor less) cards are turned upside down. In how many least number of tries can the cards be turned upside down such that all the six cards show the number ‘2’ on the upper side?

Six identical cards are placed on a table. Each card has the number ‘1’ marked on one side and the number ‘2’ marked on its other side. All the six cards are placed in such a manner that the number ‘1’ is on the upper side. In one try, exactly four (neither more nor less) cards are turned upside down. In how many least number of tries can the cards be turned upside down such that all the six cards show the number ‘2’ on the upper side? Correct Answer 3

Given:

All cards marked as 1 on One side 

And marked as 2 on Other Side

In one try, we can turn exactly 4 cards.

Calculation:

Initially:

All six cards show ‘1’

1 1 1 1 1 1

First turn:

Turn over four of the '1' cards. Now 4 cards show '2', and 2 cards show '1'.

2 2 2 2 1 1

Second turn:

Turn over three of the '2' cards and one of the '1' cards. Now 2 cards show '2' and four cards show '1'.

2 1 1 1 2 1

Third turn:

Turn over the four cards showing '1'. Now all six cards show '2'.

2 2 2 2 2 2

∴ In 3 tries we can turn all cards to number 2

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