No.
Starting with 7, every other multiple is odd.
7 multiply by 7 equal 14
For 2 to be a factor, the number has to be even. If you multiply two odd numbers together, you get an odd number. 7 is odd, so if I multiply it by any odd number, I will get a number that doesn't have 2 as a factor (e.g. 7x5=35 which doesn't have 2 as a factor). Thus, the answer to the question is "no," not all multiples of 7 have 2 as a factor.
If you are multiplying by integers, only half of them. Odd times odd is odd. Odd times even is even. Here are the products of seven and the first 10 integers (1 thru 10): 7, 14, 21, 28, 35, 42, 49, 56, 63, 70 Every other number is odd and every other number is even, depending on where you start.
105First multiply 1 and 3 to get 3. Second multiply 3 and 5 to get 15. Third multiply 15 and 7 to get 105.
Multiply 7 by any odd number, then check whether the result is between 5 and 50.
Multiply two odd numbers Add an even and an odd Subtract an odd and an even
Every number is divisible by any non-zero number. No, there are an infinite number of odd integers which are not evenly divisible by 7..
An odd number times an odd number is an odd number no matter which 2 odds you multiply.
Just multiply 7 by some odd number. Use trial and error until you find a number in the desired range.
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.
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