Let the smallest integer be 2n, the next integer is 2n + 2, and the largest one is 2n + 4. Then, we have:2n + (2n + 2) = (2n + 4) + 164n + 2 = 2n + 202n = 18 the smallest integer2n + 2 = 18 + 2 = 20 the next one2n + 4 = 18 + 4 = 22 the largest integerCheck
In general mathematicians use n to signify any integer. On this basis 2n must be an even integer.But, are you sure it was a mathematician who wrote 2n ? There's no law about this.
246 and 248.Any even number can be written in the form 2n, for some integer n. The next consecutive integer is 2n + 2. These two numbers added together equal 494:2n + (2n + 2) = 494so4n + 2 = 4944n = 492n = 123so2n = 246 and 2n +2 = 248.
The answer depends on n. If n is an integer or half on integer then 2n is a whole number; if not, then it isn't.
An odd number can be written in the form '2n+1' where 'n' is an integer, and an even number can be written int the form '2n'. We can then write the sum of 2 odd numbers as: (2n+1) + (2m+1) * Combining and factoring out a 2, we arrive at: 2(n + m + 1) Since 'n' and 'm' are both integers, we know that the value contained int the '()' is also an integer. We can therefore rewrite this equation to be: 2n Which represents an even number as given previously. * 'm' is an integer
Let the smallest integer be 2n, the next integer is 2n + 2, and the largest one is 2n + 4. Then, we have:2n + (2n + 2) = (2n + 4) + 164n + 2 = 2n + 202n = 18 the smallest integer2n + 2 = 18 + 2 = 20 the next one2n + 4 = 18 + 4 = 22 the largest integerCheck
In general mathematicians use n to signify any integer. On this basis 2n must be an even integer.But, are you sure it was a mathematician who wrote 2n ? There's no law about this.
246 and 248.Any even number can be written in the form 2n, for some integer n. The next consecutive integer is 2n + 2. These two numbers added together equal 494:2n + (2n + 2) = 494so4n + 2 = 4944n = 492n = 123so2n = 246 and 2n +2 = 248.
They are numbers of the form 2n and 2n+2 where n is any integer.
2n, where n is an integer.
Let 2n be the first integer, then the next on is 2n+2 and the one after is 2n+4 so adding all three we have 6n+6=108 or 6n=102 and we want 2n so divide we divide 6n by 3 and we have 2n=34, and 2n+2=36 and lastly 2n+4=38 so 34, 36, 38 are the even integers we seek.
Answerno solutionProcedureAssume the numbers are:2n, 2n+2, 2n+4, 2n+6, 2n+8, 2n+10Summation = 12n + 30 = 60012n = 597n = not integer numberno solution
Let's take a look at this. For any integer n, 2n always be even, then the next consecutive number 2n + 1 must be odd. Let add them first, 2n + 2n + 1 = 4n + 1 = 2(2n) + 1 So their sum is odd, since every even number multiplied by 2 is also even. Let's multiplied them, 2n(2n + 1) = (2n)^2 + 2n Their product is even, since every even number raised in the second power is also even, and the sum of two even numbers is even too. So the answer is that when the sum of two numbers can be odd, their product is an even number. (note that the sum of two odd numbers is even)
YES Even Number by definition any integer that can be divided exactly by 2. To get an even number, multiply an integer by 2 or in mathematical statement it is 2n. An even number (2x) is multiplied to any number (y) is 2x (y). So, by associative principle of multiplication 2x(y) = 2 (x*y). If (x*y) now represents any integer then 2(x*y)= 2n. And 2n is an even number.
The product of two even numbers is always an even number.Here is the proof:We define an even number as a number of the form 2n for some integer n.Now let 2n be one even number and 2m be another.The product is (2n)(2m)=2(2mn) and of course 2mn is an integer since the integers are closed under multiplication. Hence, 2(2mn) is an even number.
The answer depends on n. If n is an integer or half on integer then 2n is a whole number; if not, then it isn't.
If x is the smallest odd integer, then x = 2n + 1 for some integer n. Then the next two odd integers are 2n + 3 and 2n + 5 So the question then becomes: 2n+1 + (2n+3) + (2n+5) = 45 or 6n + 9 = 45 to be solved for n and thence the smallest of the three consecutive odd integers.