Write three numbers whose sum is 10 in all horizontal vertical and diagonal?
The sum of two numbers will almost always be greater than either number. The only exception would be when dealing with two negative numbers.
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)
3
Not always, if the smaller number is 0 or a negative number. Then their sum will be equal or less than the greater number.
Their sum is three times the middle number.
The sum of 3 consecutive whole numbers is always equal to 3 times the middle number in that sequence.
There are no three consecutive numbers with a sum of 170.
26.
That doesn't work. The number has to be divisible by three. Any three consecutive numbers add up to a multiple of three.
31 + 32 + 33 = 96
The three consecutive even numbers with a sum of 138 are 44, 46 and 48.
19, 20, 21
The three consecutive whole numbers you are looking for are 1, 2, and 3. The sum of the first two numbers, 1 + 2 = 3.
The numbers are 41, 42 and 43.
The numbers are 45, 46 and 47.
71,72,73 :)