odd numbers

If a is an even integer and b is an odd integer then prove a − b is an odd integer Let a be our even integer Let b be our odd integer We can express a = 2x (Standard form for even numbers) for some integer x We can express b = 2y + 1 (Standard form for odd numbers) for some integer y a - b =...

Let us take an integer x which is both even and odd. As an even integer, we write x in the form 2m for some integer m As an odd integer, we write x in the form 2n + 1 for some integer n Since both the even and odd integers are the same number, we set them equal to each other 2m = 2n + 1...

Take two arbitrary integers, x and y We can express the odd integer x as 2a + 1 for some integer a We can express the odd integer y as 2b + 1 for some integer b x + y = 2a + 1 + 2b + 1 x + y = 2a + 2b + 2 Factor out a 2: x + y = 2(a + b + 1) Since 2 times any integer even or odd is always...

if 2z-1 is an odd integer what is the preceding odd integer? The preceding odd integer is found by subtracting 2: 2z - 1 - 2 2z - 3

n^2+n = odd Factor n^2+n: n(n + 1) We have one of two scenarios: If n is odd, then n + 1 is even. The product of an odd and even number is an even number If n is even, then n + 1 is odd. The product of an even and odd number is an even number

n^2-n = even Factor n^2-n: n(n - 1) We have one of two scenarios: If n is odd, then n - 1 is even. The product of an odd and even number is an even number If n is even, then n - 1 is odd. The product of an even and odd number is an even number

If n represents an odd integer what represents the previous smaller odd integer Each odd integer is 2 away from the last one, so the previous smaller odd integer is found by subtracting 2 from n: n - 2

2 consecutive odd integers such that their product is 15 more than 3 times their sum. Let the first integer be n. The next odd, consecutive integer is n + 2. We are given the product is 15 more than 3 times their sum: n(n + 2) = 3(n + n + 2) + 15 Simplify each side: n^2 + 2n = 6n + 6 + 15 n^2...

The sum of 5 odd consecutive numbers is 145. Let the first odd number be n. We have the other 4 odd numbers denoted as: n + 2 n + 4 n + 6 n + 8 Add them all together n + (n + 2) + (n + 4) + (n + 6) + (n + 8) The sum of the 5 odd consecutive numbers equals 145 n + (n + 2) + (n + 4) + (n + 6)...

Roberto has taken 17 photos photos are placed on each odd number page and the newspaper has 10 pages total. The pages with photographs will have 3 or 4 photos each. How many pages has 3 photos and how many pages have 4 photos? Odd pages are 1, 3, 5, 7, 9 17/5 = 3 with 2 remaining. So all 5...