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The product of two consecutive integers is 121 Find the numbers?
As there are two consecutive integers then one must be an even number and the other an odd number. If the numbers are y and y + 1 then if y is even, y + 1 is odd and if y is odd then y + 1 is even.The product of an even integer and an odd integer is always even.The question therefore has no answer.121 = 112 but this is not what the question has asked.
Are all odd square roots the square roots of odd numbers?
Yes, they are. Sketch of proof: Odd square root is a number x of form x = 2n + 1. Square of this root is: x2=(2n+1)2=4n2+4n+1=y, which can be expressed as: y=2(2n2+2n)+1. Let t = 2n2+2n. Then, y=2t+1, which is an odd number.
What are the 5 odd numbers whose sum is 50 from 1 to 19?
It is not possible for five odd numbers to add up to 50 or any even number. All even numbers have the form 2x, where x is an integer, that is if you divide an even number by two you get an integer. All odd numbers have the form 2x + 1, where x is an integer. Another way of looking at this is that every odd number is an even number plus 1. So lets take five odd numbers, 2v+1, 2w+1, 2x+1, 2y+1, and 2z+1, where v, w, x, y, and z are distinct integers. Adding them together we get: (2v+1) + (2w+1) + (2x+1) + (2y+1) + (2z+1) 2v + 2w + 2x + 2y + 2z + 5 2v + 2w + 2x + 2y + 2z + 4 + 1 2(v + w + x + y + z + 1) +1 This is the form of an odd number. Therefore any five odd numbers added together will sum to an odd number.
What are the factors of xy plus x plus y plus 1?
xy + x + y + 1 = (x + 1)(y + 1).
Why do two even numbers always equal an even number?
On a number line, adding an even number to another number (or zero) results in an even displacement, which must end in the same type of number as the original. If the beginning number is odd, adding an even number produces an odd sum. If the beginning number is even, adding an even number produces an even number. A corollary is that: Adding two like numbers produces an even number. Adding two unlike numbers produces an odd number. ---------------------------------- Algebraically: let x be an even number, Then y = 2x for some value of x Two even numbers would be 2m and 2n Adding them gives: 2m + 2n = 2(m + n) = 2p where p = m + n; 2p is of the form y = 2x, so 2p is an even number. Thus adding two even numbers results in an even number. Similarly for odd numbers: If y = 2x is an even number then z = y + 1 = 2x +1 is an odd number. Two odd numbers would be 2m+1 and 2n+1 Adding them gives: 2m+1 + 2n + 1 = 2m + 2n + 2 = 2(m + n + 1) = 2p where p = m + n + 1 Thus adding two odd numbers results in an even number. Similarly for one even and one odd number. An even number would be 2m and an odd number would be 2n+1 Adding them gives: 2m + 2n + 1 = 2(m + n) + 1 = 2p + 1 where p = m + n Thus adding an even number and an odd number results in an odd number.
The sum of two odd numbers is always odd?
The sum of any two odd numbers is always even. This is because (a) any odd number can always be expressed as an even number plus 1, and (b) the sum of two even numbers is always even. So if X is even and Y is even then X+1 added to Y+1 will equal X+Y+2
The product of two consecutive integers is 121 Find the numbers?
As there are two consecutive integers then one must be an even number and the other an odd number. If the numbers are y and y + 1 then if y is even, y + 1 is odd and if y is odd then y + 1 is even.The product of an even integer and an odd integer is always even.The question therefore has no answer.121 = 112 but this is not what the question has asked.
Are all odd square roots the square roots of odd numbers?
Yes, they are. Sketch of proof: Odd square root is a number x of form x = 2n + 1. Square of this root is: x2=(2n+1)2=4n2+4n+1=y, which can be expressed as: y=2(2n2+2n)+1. Let t = 2n2+2n. Then, y=2t+1, which is an odd number.
When you add an odd and an even number together is the sum odd or even Why?
The result is odd. An odd number leaves a remainder of 1 when it is divided by 2 so it may be written as 2x + 1 where x is an integer. An even number must be a multiple of 2 so it may be written as 2y where y is an integer. The sum of these two is 2x + 1 + 2y = 2*(x + y) + 1 That is to say that if it is divided by 2, you get a quotient of x+y and a remainder of 1 - ie it is odd.
What is -5x plus 2y equals 29?
An equation with an infinite number of solutions eg: x = 1, y = 17, x = 17, y = 57, x = -3, y = 4 Solutions for y will be in whole numbers when x is odd.
Find the critical number of y-1 divide by y2-y plus 1?
y-1/y2-y+1=y/y-1
Why when you add odd and even numbers the sum is odd?
Factors can be listed in pairs. Square numbers will have one pair that is the same number. When put into a list, that number will only be listed once and will result in an odd number of factors.
Why does an even number subtract to odd number is equal to odd number?
Suppose x is an odd number, then x leaves a remainder when divided by 2. That is, x = 2m+1 (for some integer m). Suppose y is an even number, then y is a multiple of 2 so suppose y = 2n (for some integer n). Then x-y = 2m+1 - 2n = 2m-2n + 1 =2(m-n) + 1 Since m and n are integers, then m+n is an integer so that the sum gives 1 more than a multiple of 2. And that is what an odd number is!
How would you prove that the sum of an even number and an odd number is always an odd number?
Suppose x is an even number and y is an odd number. Then x = 2*n for some integer n and y = 2*m + 1 for some integer m Therefore x + y = 2*n + 2*m + 1 = 2*(n + m) +1 Now, since n and m are integers, (n + m) is also an integer [by the closure of integers under addition]. Thus, x + y = 2*p + 1 where p = n + m is an integer. ie x + y is an odd integer.
What are the 5 odd numbers whose sum is 50 from 1 to 19?
It is not possible for five odd numbers to add up to 50 or any even number. All even numbers have the form 2x, where x is an integer, that is if you divide an even number by two you get an integer. All odd numbers have the form 2x + 1, where x is an integer. Another way of looking at this is that every odd number is an even number plus 1. So lets take five odd numbers, 2v+1, 2w+1, 2x+1, 2y+1, and 2z+1, where v, w, x, y, and z are distinct integers. Adding them together we get: (2v+1) + (2w+1) + (2x+1) + (2y+1) + (2z+1) 2v + 2w + 2x + 2y + 2z + 5 2v + 2w + 2x + 2y + 2z + 4 + 1 2(v + w + x + y + z + 1) +1 This is the form of an odd number. Therefore any five odd numbers added together will sum to an odd number.
If you add2 odd numbersis the answer odd or even?
Answer is even,to find the answer just pick numbersOdd + Odd = Even3 + 3 = 6Mathematical explanationA odd number can be written 2*x+1, where x is any integer, and an even can be written 2x.2x+1 = 2y+1 = 2(x+y) +2 = 2z, where z=x+y+1
Why is the sum of three odd numbers odd?
Let x, y and z be even numbers. This means that (x + 1), (y + 1) and (z + 1) are all odd. Adding these three odd numbers gives (x + y + z + 3). Because the sum of any number of evens must be even as well - which can be easily proven with idea that where a = 2m and b = 2n, a + b = 2m + 2n = 2(m + n), and is even - adding 3 changes it to an odd number. This holds for any set of three odd numbers. i agree with this... but what about in the form of 2n+1?
