21, 21, 21, 21 , 15, 1
The numbers are 11, 13 and 15.
There are 6 odd numbers that are greater than 6 but less than 19. The odd numbers between 6 and 19 are 7, 9, 11, 13, 15, and 17.
The first six integers (0,1,2,3,4,5) add up to 15, unless you exclude 0, in which case the answer is 21.
Never. When you dont add the same odd number and even number of times. For example: 5 + 3 (2 odd numbers; and even amount of odd numbers ) = 8 5 + 3 + 7 = 15 an odd amount of odd numbers 5 + 3 + 7 + 9 = 24 another even amount of odd numbers. But never if you use the only same number
21, 21, 21, 21 , 15, 1
The sum of six odd numbers is never fifteen since 15 is odd and the sum of six even numbers is always even. However if "make" allows products: 1* 1 * 1 * 1 * 3* 5 = 15 ( I used * to mean multiply.)
15
The numbers are 11, 13 and 15.
No, because two odd numbers always add together to be an even number, and a even plus an odd number is an odd number, example: 3+1+11=An even number? 3+1=4 4+11=15 Is 15 even? No. 3+1+11= An odd number (15)
You can add any one of 13, 15, 17, 19 and 21.
15, 17, 19, 21 and 23.
You cannot add 4 odd NUMBERS to make an odd number, so you have to think constructively...eg 1 + 3 + 15 = 19. There are other similar solutions...
You cannot do this, because you're taking odd numbers: if you add two odd numbers together, you will get an even number. Take the odd numbers in pairs, so you'll have 3 pairs plus an extra odd number. The 3 pairs make 3 even numbers. You can add up any number of even numbers and get an even number, so you will have an even + odd, which is odd. 30 is an even number, but your result will be odd, no matter what 7 odd numbers you choose.
There are 6 odd numbers that are greater than 6 but less than 19. The odd numbers between 6 and 19 are 7, 9, 11, 13, 15, and 17.
The first six integers (0,1,2,3,4,5) add up to 15, unless you exclude 0, in which case the answer is 21.
Yes. All numbers that do not give a decimal-free number when divided by 2 are odd numbers. We get 15/2=7.5 => 15 is an odd number;)