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Q: Is answer even or odd when you multiply odd x even x even?

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Well, the question is why. The first number is "even" + 1. Multiply both of these by your odd number. Now the "even" times "odd" is even, because every "1" in the odd number becomes a "2". And then the remaining 1 times "odd" must be odd, which is an even +1. Add it all up and you get evens everywhere except that final "1". So the result is even + 1 which is odd. There is a quicker way if you know how to multiply bracketed terms: odd x odd = (even + 1)x(even +1)= even x even +even +even +1 = must be odd. ========================== You've just read a truly impressive answer to a question slightly different from the one that was asked. The part of the question that comes after "Why if ..." is a false statement. If you multiply odd number with another number, the result is odd number ONLY if the nother number is also odd number.

Multiply two odd numbers Add an even and an odd Subtract an odd and an even

because that's the theory of maths even x even = even even x odd = odd even + even = even even + odd = odd

Yep

You get an even number as your answer.

Related questions

If you multiply an odd number by an even number, the product (answer) will be even. Example: 5 x 4 = 20

If you multiply one even number by one odd number, the result is always even. In general, if you multiply several numbers, and at least one of the numbers is even, the product is always even. This is because "even" means "multiple of 2", and if one of the factors contains a 2 as a factor, so will the product.

An even number. For example: 3 (odd) x 2 (even) x 6 (even) = 36 (even) Second example: 5 (odd) x 8 (even) x 2 (even) = 80 (even)

yes, the product of 2 odd numbers is always an odd number. Well, the question is why. The first number is "even" + 1. Multiply both of these by your odd number. Now the "even" times "odd" is even, because every "1" in the odd number becomes a "2". And then the remaining 1 times "odd" must be odd, which is an even +1. Add it all up and you get evens everywhere except that final "1". So the result is even + 1 which is odd. There is a quicker way if you know how to multiply bracketed terms: odd x odd = (even + 1)x(even +1)= even x even +even +even +1 = must be odd.

Well, the question is why. The first number is "even" + 1. Multiply both of these by your odd number. Now the "even" times "odd" is even, because every "1" in the odd number becomes a "2". And then the remaining 1 times "odd" must be odd, which is an even +1. Add it all up and you get evens everywhere except that final "1". So the result is even + 1 which is odd. There is a quicker way if you know how to multiply bracketed terms: odd x odd = (even + 1)x(even +1)= even x even +even +even +1 = must be odd. ========================== You've just read a truly impressive answer to a question slightly different from the one that was asked. The part of the question that comes after "Why if ..." is a false statement. If you multiply odd number with another number, the result is odd number ONLY if the nother number is also odd number.

Multiply two odd numbers Add an even and an odd Subtract an odd and an even

No. If the number by which you multiply 5 is even, the product will always end in "0" and will be even:5 x 2 = 105 x 8 = 405 x 18 = 90If the number by which you multiply 5 is odd, the product will always end in "5" and will be odd:5 x 3 = 155 x 9 = 455 x 31 = 155

For 'odd' prime numbers this is true, but '2' is also a prime number , so you will have an even answer. Remember when multiplying even X even = even even X odd = even odd X even = even odd x odd = odd.

Odd times even is even.

No, only every other multiple is odd. Example: 3,6,9,12,15,18

because that's the theory of maths even x even = even even x odd = odd even + even = even even + odd = odd

The multiples of all odd numbers are odd and even. Odd x odd = odd. Odd x even = even. Since odd and even numbers alternate, the multiples will alternate as well.