No.1, 3, and 5 are odd numbers.
If your only option is to assemble the digits into a single number, then the answer is 0, as you need at least one even digit to make an even number, and the digits provided are all odd ones. If on the other hand, you can use operators on them (e.g. 1 + 3 + 5 + 7), then you actually have quite a large number of possibilities. For example: 1 + 3 + 5 + 7 - 1 + 3 + 5 + 7 1 - 3 + 5 + 7 (1 + 3) / ( 7 - 5) 1 + √(7 + 5 - 3) 31 + 57 etc. etc. Alternatively, if you're not worried about them being even numbers, then the answer is 24.
In order for the number to be even, the '2' must be the last digit. So we only need tofigure out how many ways there are to arrange the 1, 3, and 5 in the first 3 places.The first place can be any one of 3 digits. For each of those ...The second place can be either one of the 2 remaining digits.And then the third place has to be the only digit that's left.So there are (3 x 2) = 6 ways to arrange 1, 3, and 5 in the first 3 places.That means only 6 even 4-digit numbers can be made from the digits 1, 2, 3, and 5.And here they are:1 3 5 21 5 3 23 1 5 23 5 1 25 1 3 25 3 1 2
50%, 1/2 or 3/6. 1, 3 and 5 are odd, and 2, 4 and 6 are even.
if the numbers are from 1 - 9 then the ratio of even to odd is 4:5 (2, 4, 6 & 8 are even, 1, 3, 5, 7 & 9 are odd)
Numbers that end with 0, 2, 4, 6, or 8 are even. Numbers that end with 1, 3, 5, 7, or 9 are odd.Numbers that end with 0, 2, 4, 6, or 8 are even. Numbers that end with 1, 3, 5, 7, or 9 are odd.Numbers that end with 0, 2, 4, 6, or 8 are even. Numbers that end with 1, 3, 5, 7, or 9 are odd.Numbers that end with 0, 2, 4, 6, or 8 are even. Numbers that end with 1, 3, 5, 7, or 9 are odd.
5/(√3 - 1)= 5(√3 + 1)/(√3 - 1)(√3 + 1)= (5√3 + 5)/[(√3)2 - 12)= (5√3 + 5)/(3 - 1)= 5√3 + 5)/2= 5√3/2 + 1/2
6
to be an even multiple of 3 and 5 means it has to be a multiple of 2x3x5 = 30 answer is 5 numbers: 30, 60, 90, 120, 150
To divide by a fraction, multiply by its reciprocal. In this instance, 3 / 1/5 = 3 x 5 = 15
there are none
odd endings 1 3 5 7 9
There is no solution to this problem. If each digit can be used once only then we have 5 odd numbered digits (1,3,5,7,9) and 4 even numbered digits (2,4,6,8). To create the two numbers that are added together requires the following combinations of digits. 5 Odd & 1 Even ....when added these will generate 2 Even digits & 1 Odd digit but the remaining digits are 3 Even. 4 Odd & 2 Even. These will generate 3 Even digits OR 1 Even digit & 2 Odd digits but the remaining digits are 1 Odd & 2 Even. 3 Odd & 3 Even. These generate 3 Odd digits OR 2 Even & 1 Odd digits but the remaining digits are 2 Odd & 1 Even. 2 Odd & 4 Even. These generate 3 Even digits OR 2 Odd & 1 Even digits but the remaining digits are 3 Odd & no Even.