Happy new year everyone.
My first quiz for this year
What is the smallest number of coins that need to be moved so that the coins inside one box total exactly twice the value of those in the other? Think laterally for this one!
100% upvote for the first correct answer :-)
,
One move, you put the box with the nine in it inside the other box which already has a value of nine (1+2+2+4=9). Then the box you moved has 9 and the other box has 18(1+2+2+4+9=18), which is twice the value. Is that right?!
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well done :-) how much time take you to have this answer :-)
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Omg it really was hard. I sat there for 10 minutes wondering. I usually don't have that much patience to figure out a question but I really wanted to get it. Thanks for sharing this. Please continue to post challenging riddles.
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Coin 1 and 2 from the first box is transferred to the second box leaving coin 2 and 4 in the first box. 2 + 4 from the first box equals 6 which is half the summation of coins in the second box. Thank you
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No, think more :-)
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For you to get that answer,I think that question needs to be rephrased.
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you have above a correct answer
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2, if you move the 1 and the 2 over to the box with the 9 then the left box totals 6 and the right box totals 12.
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No :-)
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move the 1 from the first box to make 16 in the second box, 16 is twice 8
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No, try more options !
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move the 2 over from the first box, making 124 in the first box and 62 in the second box
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You need to look for the smallest number of coins which should be moved ...to have the result :-)...in the second box is 9 not 6
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