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How do you make multiples of 6 using 0-9 only once? - Answers

Here are two solutions: 1234567890 and 9876543210. There are many more.To be a multiple of 6, the number must be both a multiple of 3 and a multiple of 2. If the sum of the digits is a multiple of 3, then the number is a multiple of 3. Adding up all digits (0 through 9) gives a sum of 45, which is a multiple of 3. Since addition is commutative, rearranging the digits in any order will still have a sum of 45, so it's still a multiple of 3. For it to be a multiple of 2, the last digit (the one's place) must be {0,2,4,6, or 8}



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How do you make multiples of 6 using 0-9 only once? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_make_multiples_of_6_using_0-9_only_once

Here are two solutions: 1234567890 and 9876543210. There are many more.To be a multiple of 6, the number must be both a multiple of 3 and a multiple of 2. If the sum of the digits is a multiple of 3, then the number is a multiple of 3. Adding up all digits (0 through 9) gives a sum of 45, which is a multiple of 3. Since addition is commutative, rearranging the digits in any order will still have a sum of 45, so it's still a multiple of 3. For it to be a multiple of 2, the last digit (the one's place) must be {0,2,4,6, or 8}



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https://math.answers.com/math-and-arithmetic/How_do_you_make_multiples_of_6_using_0-9_only_once

How do you make multiples of 6 using 0-9 only once? - Answers

Here are two solutions: 1234567890 and 9876543210. There are many more.To be a multiple of 6, the number must be both a multiple of 3 and a multiple of 2. If the sum of the digits is a multiple of 3, then the number is a multiple of 3. Adding up all digits (0 through 9) gives a sum of 45, which is a multiple of 3. Since addition is commutative, rearranging the digits in any order will still have a sum of 45, so it's still a multiple of 3. For it to be a multiple of 2, the last digit (the one's place) must be {0,2,4,6, or 8}

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      Here are two solutions: 1234567890 and 9876543210. There are many more.To be a multiple of 6, the number must be both a multiple of 3 and a multiple of 2. If the sum of the digits is a multiple of 3, then the number is a multiple of 3. Adding up all digits (0 through 9) gives a sum of 45, which is a multiple of 3. Since addition is commutative, rearranging the digits in any order will still have a sum of 45, so it's still a multiple of 3. For it to be a multiple of 2, the last digit (the one's place) must be {0,2,4,6, or 8}

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