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What are two consecutive odd integers whose sum is 56?
1. let n = the odd integer then n + 2 must be the next odd integer 2. n + (n + 2) = 56 3. 2n + 2 = 56 4. 2n = 54 5. n = 27 6. 27 + 2 = 29 is the next odd integer
Is the sum of an even whole number and an odd whole number always odd?
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
When you multiply an odd number and an even number is your answer always odd?
No, it's always even, and here's the proof: All even numbers can be expressed as 2n, where n is any integer. All odd numbers can be expressed as 2p + 1, where p is any integer. Multiply those two together: 2n(2p + 1) = 2(2np + n). Since both 2np and n are integers, that means 2np + n is an integer; and since that integer is being multiplied by 2, it must be even.
If 2n is an even integer what is the next even integer?
2n + 2 = 2(n+1)
Why is zero an even number?
Why Zero is EvenDivision : By definition, an even number is divisible by 2, with no remainder. 0/2 = 0 with no remainder.Number Line : on the integer number line, even and odd numbers alternate, and any odd number increased or reduced by 1 is an even number.Addition : additive rules state that even + even = even, and odd + odd = even. Under these rules, zero can be even, but cannot be odd.With the obvious solution n=0, zero fits the criteria that for any integer n, the integer 2n is even. (multiplication or division by 2)
If n plus 4 represents an odd integer the next larger number odd integer is represented by?
Every integer is either even (divisible by 2) or odd (not divisible by 2). Since an even number plus 1 is odd and an odd number plus one is even, because 1 does not divide 2. We know (n + 4) is odd. The next integer is (n + 4 + 1) = (n + 5), because an odd number plus 1 is even, (n + 5) is even. The integer after (n + 5) is (n + 6), since (n + 5) we know is even, (n + 6) must be odd. Since (n + 6) is the smallest integer that is greater than (n + 4) and is odd, so (n + 6) is the next odd integer.
What are the next two consecutive odd integers if n is an odd integer?
If n is an odd integer then the next two consecutive odd integers are n+2 and n+4.
If n is an odd integer is n plus 6 odd as well?
Yes. If n is odd, then n + c where c is an even constant will be odd. n + d where d is an odd constant will be even.
What are two consecutive odd integers whose sum is 56?
1. let n = the odd integer then n + 2 must be the next odd integer 2. n + (n + 2) = 56 3. 2n + 2 = 56 4. 2n = 54 5. n = 27 6. 27 + 2 = 29 is the next odd integer
If N-3 is an even integer what is the next larger consecutive even integer?
n-3
Why is 1 an odd number?
Odd numbers are defined as numbers of the form 2n+1 for some integer n. Since 0 is an integer, 2x0+1=1 is an odd number. Another way to think of it is that an odd number is one that is not even. An even number is one of the form 2n for some integer n. 1 is not even so it is odd. This means 0 is even, by the way.
Why is the sum of an even and odd number alsways odd?
A number a is even if there exists an integer n such that a = 2n A number b is odd if there exists an integer m such that b = 2m + 1. So: a+b = (2n) + (2m +1) = 2 (n+m) + 1 Since n and m are integers, n+m is also an integer. So a+b satisfies the definition of an odd number.
What is an expression for 'the next integer'?
If your integer is "n", then the next integer will be "n+1".
Can the sum of an odd and even number be odd?
Yes, 1+2=3 and since 1 is odd and 2 is even this shows it can happen. In fact it always does! If we let an even number be denoted by 2n for some integer n and and an odd one 2m+1 for some integer m. Then, 2m+1+2n=2(m+n)+1. Call m+n, p since it is also an integer. 2p+1 is always odd.
Is the sum of an even whole number and an odd whole number always odd?
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
When you multiply an odd number and an even number is your answer always odd?
No, it's always even, and here's the proof: All even numbers can be expressed as 2n, where n is any integer. All odd numbers can be expressed as 2p + 1, where p is any integer. Multiply those two together: 2n(2p + 1) = 2(2np + n). Since both 2np and n are integers, that means 2np + n is an integer; and since that integer is being multiplied by 2, it must be even.
What does consecutive even integers mean?
An even integer is a number that is a multiple of 2. If n is an even integer, the next consecutive even integers are n+2, n+4 and so on.
