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A two-digit integer is written next to itself twice, forming a six-digit number. If the resulting number is divisible by 6, then how many possibilities are there for the original two-digit number?

Guest Nov 4, 2018
edited by Guest  Nov 4, 2018
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First, create six spots for your number and label them 100000x, 10000y, 1000z, 100x, 10y, z. Add everything up, and see what you can do from there! 

ant101  Nov 4, 2018
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I'm not sure about this, but I think it always must be divisible by 3, so all we need is for the original to be divisible by 2. So there are 9 digits for the tens place of the original and 5 places for the ones digit, so 45. Can someone check this?

Guest Nov 5, 2018

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