199 The first odd number is 1 The second odd number is 3 The rth odd number is thus 2r - 1 Thus the 100th odd number is 2 x 100 - 1 = 199.
Let one odd number be "2m + 1", the other odd number "2n + 1" (where "m" and "n" are integers). All odd numbers have this form. If you multiply this out, you get (2m+1)(2n+1) = 4mn + 2m + 2n + 1. Since each of the first three parts is even, the "+1" at the ends converts the result into an odd number.
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.
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.
There is a surprisingly easy formula for this. Sum of n odd numbers = n2 So the sum of the first 600 odd numbers (starting with 1 as the very first odd number) is 6002 = 360000.
Answer:1 is an odd number If you need to learn math go to the web site in the link below
199 The first odd number is 1 The second odd number is 3 The rth odd number is thus 2r - 1 Thus the 100th odd number is 2 x 100 - 1 = 199.
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The first prime number that is odd is 3, so the second odd prime number is 5. Some think 1 is the first prime nunber, but 1 is neither prime nor composite.
The first odd prime number is 3, not 1, because 1 has only 1 factor, 1, while prime numbers must have 2 factors, 1 and itself.
Let one odd number be "2m + 1", the other odd number "2n + 1" (where "m" and "n" are integers). All odd numbers have this form. If you multiply this out, you get (2m+1)(2n+1) = 4mn + 2m + 2n + 1. Since each of the first three parts is even, the "+1" at the ends converts the result into an odd number.
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.
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.
-- The first odd number is some even number plus an extra ' 1 '. -- The second odd number is some even number plus an extra ' 1 '. -- Add them. You get (an even number) + (an even number) + ( 2 ). That's an even number. ======================== Another mind-bending , brain-busting way to look at it : -- The first odd number is some even number plus an extra ' 1 '. -- The second odd number is some even number minus ' 1 '. -- Add them. You get (an even number) + (an even number) + ( [1 - 1] or zero ). That's an even number.
The number 1 is an odd number.
Here's the following proof that "product of two odd numbers is odd". Proof: Any odd number can be written in the form 2p+1. Let your first odd number be written in this form. Let your second odd number be written as 2q+1 - essentially the same form as above, but since p may not equal q, separate variables are used. Thus: First odd number x second odd number = (2p+1)(2q+1) = 2pq + 2p +2q +1. The plus one on the end indicates that the product is odd, as required.
There is a surprisingly easy formula for this. Sum of n odd numbers = n2 So the sum of the first 600 odd numbers (starting with 1 as the very first odd number) is 6002 = 360000.