0
This is a bit of a funny one, while some may argue that 0 is neither odd or even and a bit of a special integer in the way 1 is neither prime nor composite, but when you look at it logically in terms of integers and the definition of odd and even, 0 is an even number.
By definition an even integer is any integer that can be expressed as 2k, where the k is any number is the set of integers (all the whole numbers from -infinity all the way to +infinity). In other words an even number is any integer that can be divided by 2 and still leave an integer, and it is quite apparent that 0 fits this qualification, as 0/2 = 0
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Are the multiplier of even number always even?
In a mathematical context, a multiplier for a number, r, is be (1 + r/100) which is usually a rational fraction and the concept of odd or even does not apply to fractions.
How would you explain that the product of any two consecutive number is even?
If we take two consecutive numbers then one of them is always even and other is odd.And product of an even and an odd number is always even.Reason:Let us say the even number be m and the odd number be n.The even number m can be expressed as 2 x p, where p is some natural number.Now, a thing to remember!"For two natural numbers q and r, q is divisible by r if we can express q as r x s where s is some natural number."Product of m and n = m x n = 2 x p x n.It can be seen that the result is divisible by 2. And if a product is divisible by 2 then it means the product is even.
A c program to find maximum number in a 3 by 3 matrix?
int matrix[][]; // the matrix to find the max in int max = matrix[0][0]; int r,c; for(r = 0; r < 3; ++r) { for(c = 0; c < 3; ++c) { if(matrix[r][c] > max) { max = matrix[r][c]; } } } // max is now the maximum number in matrix
What does binary code mean?
170/2 = 85 R 0. Therefore binary number so far is 0.85/2 = 42 R 1. Therefore binary number so far is 10.42/2 = 21 R 0. Therefore binary number so far is 010.21/2 = 10 R 1. Therefore binary number so far is 1010.10/2 = 5 R 0. Therefore binary number so far is 01010.5/2 = 2 R 1. Therefore binary number so far is 101010.2/2 = 1 R 0. Therefore binary number so far is 0101010.1/2 = 0 R 1. Therefore binary number so far is 10101010.The integer portion of last division was 0 so for the decimal number 170, the binary equivalent is 10101010.
The product of two consecutive odd integers is 2 less than six times their sum can you find the two integers?
there cannot be any nos like that.. product of 2 odd nos is odd.. sum of two even nos is even.. that multiplied by six is even too.. subtracting 2 from that also gives an even no.. let x and y be the odd integers. according to the given question xy=6(x+y)-2..here we r actually equatin an odd no and an even no.. which is wrong.. so there cant be any two consecutive odd nos that fit in the question
Any integer is input through the keyboard write a c program to find out whether it is an odd number or even number?
printf(\n "ENTER THE NUMBER\t"); scanf{"%d",&a); while(a!=0); { r=a%2; if(r==0) printf("\n\n THE NUMBER IS EVEN \t"); else printf("\n\n THE NUMBER IS ODD \t"); printf ("\n ENTER THE NUMBER \t"); scanf("%d",&a); } getch(); }
Are the multiplier of even number always even?
In a mathematical context, a multiplier for a number, r, is be (1 + r/100) which is usually a rational fraction and the concept of odd or even does not apply to fractions.
Write a Shell program to find the area and circumference of a circle?
echo "Enter the Number" read n r=`expr $n % 2` if [ $r -eq 0 ] then echo "$n is Even number" else echo "$n is Odd number" fi
C program to find the sum of odd positioned digits and even positioned digits of a number seperately?
#include<stdio.h> #include<conio.h> main() { int n,s,r,t; clrscr(); printf("enter n"); scanf("%d",&n); s=0;t=0; while(n!=0) { r=n%10; { if(r%2!=0) t=t+r; if(r%2==0) s=s+r; } n=n/10; } printf("sum of even position digits%d\n",s); printf("sum of odd position digits%d\n",t); getch(); }
Does adding an even and odd number make a sum even?
ofcourse not.Because if we take an example of an even and odd number.Let us take a simple even number that is 2 and a simple odd number that is 3.According to the question if we add this two numbers it is not necessary that we willget the sum as even.as shown below2+3=5;From the above we can observe that we r not getting the result as even.Thus It is notnecessary that by adding an even and odd we will get the sum as even.== ==
Why does an even number plus an odd number always equal an odd number?
For any even number n, the following integer is odd (n+1), so for any odd number other than 1, you are adding the preceding even number, plus 1. Because all even numbers are multiples of 2, their sum is even, and 1 more will be an odd number.---Think of the odd number as an even number plus one and remember that all even numbers are multiples of two.Therefore:- Any two even numbers added together will always be a multiple of two and therefore even. Any two odd numbers added together will be the same as adding two even numbers plus 1 plus 1. (A multiple of two plus two which is still a multiple of two).An odd number added to an even number will be the same as two even numbers plus 1 in other words a multiple of two (even) plus one making the result odd.Extrapolating further it's easy to see that by the above reasoning that for all positive integers...Any number of even numbers added together will always be even.An odd number of odd numbers added together will always be odd.Any even number of odd numbers added together will always be even,Adding any number of even numbers to any number of odd numbers will make no difference to the outcome, the odd or even outcome will still be determined by the number of odd numbers being added together.---Another approach:Let n, r be real numbers even number representation: 2nodd number representation: 2n + 1sum = 2n + 2n + 1 = 4n + 1 = 2(2n) + 1let 2n = rsum representation becomes: 2r + 1, which is a representation of an odd number.
Is it true that the sum of two sequential prime number number is not prime?
We know that a prime number is a positive integer greater than 1, whose divisors are 1 and itself. We know that the only even prime number is 2. That means that all other prime numbers are odd numbers.We know that when we add two odd numbers the result is an even number, which are not prime numbers (expect 2, and 2 = 1 + 1 where 1 is odd but is neither prime nor composite). Thus adding two odd prime numbers cannot give us another prime number.We show that the conclusion follows from the premise:Assume that r = p + q where all r, p and q are prime numbers, then we have that r is either even or not:If r is even then r is at least 4 (the smallest number which is the sum of two primes) and thus not a prime number. This contradicts the assumption that r is a prime number, and therefore we conclude that r is not even.If r is odd then either p or q must be odd and the other one must be even, since both p and q are prime numbers one of them must be 2 (the only even prime number).2 + 3 = 5, 2 + 17 = 19, are examples of such numbers.
How do you prove that if the sum of two prime numbers is prime then one of the numbers is 2?
We know that a prime number is a positive integer greater than 1, whose divisors are 1 and itself. We know that the only even prime number is 2. That means that all other prime numbers are odd numbers.We know that when we add two odd numbers the result is an even number, which are not prime numbers (expect 2, and 2 = 1 + 1 where 1 is odd but is neither prime nor composite). Thus adding two odd prime numbers cannot give us another prime number. We show that the conclusion follows from the premise:Assume that r = p + q where all r, p and q are prime numbers, then we have that ris either even or not:* If r is even then r is at least 4 (the smallest number which is the sum of two primes) and thus not a prime number. This contradicts the assumption that r is a prime number, and therefore we conclude that r is not even. * If ris odd then either p or q must be odd and the other one must be even, since both p and q are prime numbers one of them must be 2 (the only even prime number). 2 + 3 = 5, 2 + 17 = 19, are examples of such numbers. See http://en.wikipedia.org/wiki/Twin_prime for more.
How would you explain that the product of any two consecutive number is even?
If we take two consecutive numbers then one of them is always even and other is odd.And product of an even and an odd number is always even.Reason:Let us say the even number be m and the odd number be n.The even number m can be expressed as 2 x p, where p is some natural number.Now, a thing to remember!"For two natural numbers q and r, q is divisible by r if we can express q as r x s where s is some natural number."Product of m and n = m x n = 2 x p x n.It can be seen that the result is divisible by 2. And if a product is divisible by 2 then it means the product is even.
An odd multiple of 6 which is less than 100?
All multiples of 6 are even numbers.
Draw a flowchart to find the even numbers from 1 to 100?
To draw a flowchart to find even numbers from 1 to 100, begin with a box labeled start. Assign a color to even numbers and a color to odd numbers. Beginning at the start box, make an arrow down and insert the number "1" into the box, continue adding arrows and numbers until you reach 100. If you used pink for even numbers and blue for odd numbers, each number in the list that is divisible by 2 will be colored in pink, and all the rest will be colored in blue.
How many subsets containing an odd number of elements does a set with 10 elements have?
We are looking for all sets with an odd number of elements in them. To find this we can use combinations. Adding up the odd combinations will total the subsets containing an odd number of elements.nCr where n=10 and r=(1,3,5,7,9)10C1 + 10C3 + 10C5 + 10C7 + 10C9 = 512.You could also take the total number of elements (2^10) and subtract all sets containing an even number of elements.nCr where n=10 and r=(0,2,4,6,8,10)1024-(1+45+210+210+45+1)=512
