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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.
