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.
No. The sum of 1 odd number and 1 even number must always be odd.
Odd. First subtract 1 less than your odd number, (which is obviously even) to get even - even, which is even. Then take off your spare 1 to finish up odd.
0 is a number to divide the positives and the negitives. Then it will be an even in this timeline. -2, -1, 0, 1, 2 Because it goes even odd even odd even. If zero was not there, then it would go even odd odd even.
it is a odd number it ends in 1. odd numbers end in 1 even numbers end in 2
because... odd+odd=even even+odd=odd e.g 1+1+1=3 odd+odd+odd=odd
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.
No. The sum of 1 odd number and 1 even number must always be odd.
The ratio is 1:2. For every even number there are 2 odd numbers. Fibonacci Series: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... which is even, odd, odd, even, odd, odd, even, odd, odd, even, ... the pattern continues
Odd. First subtract 1 less than your odd number, (which is obviously even) to get even - even, which is even. Then take off your spare 1 to finish up odd.
odd
0 is a number to divide the positives and the negitives. Then it will be an even in this timeline. -2, -1, 0, 1, 2 Because it goes even odd even odd even. If zero was not there, then it would go even odd odd even.
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.
*1 is an odd number * 0 is an even number
it is a odd number it ends in 1. odd numbers end in 1 even numbers end in 2
opposite - answer will always be odd (when adding odd + even) odd + even 5 + 4 = 9 odd even + even 2 + 6 = 8 even odd + odd 3 + 1 = 4 even
The two dice can be (odd, odd), (odd, even), (even, odd), or (even, even). Thus the probability of two dice totaling an even number is 2*(1/2)*(1/2) = 1/2. 50/50