This is a weekly question posted by Columbus State University. See http://www.colstate.edu/mathcontest/problem.php?CategoryID=3&LinkID=Current
Please do not answer this question as this person is too lazy to figure it out themselves.
To find the sum of binary numbers 1101 and 1110, we perform binary addition. Starting from the rightmost bit, we add each pair of bits along with the carry from the previous addition. 1+0 gives 1, 0+1 gives 1, 1+1 gives 0 with a carry of 1, and 1+1 (including the carry) gives 1 with a carry of 1. Therefore, the sum of 1101 and 1110 in binary is 11011.
First make the number line by drawing a horizontal line with an arrow at each end. Then put 7 or more evenly-spaced, vertical hatch marks ( very short vertical lines that extend about 1/8" above and below the horizontal line). Below the rightmost hatch mark put 0 (zero) , and then put the 6 negative numbers below the hatch marks to the left of zero. Put dots on the number line where the hatch marks for -5 and -6 cross it.
In Decimal: 10101 - 01110 = 8991 In Binary: 10101 - 01110 = 111 In C: 10101 - 01110 = 9517 = 022455 = 0x252d
The answer is 110111 The step is given
In order for a number to be divisible by 4, it has to be even.As every multiple of 100 is divisible by 4, we only have to look at the rightmost two digits.If the number consisting of these digits is divisible by 4, then the original number is also divisible by 4.For example:353 is odd, therefore is not divisible by 4.1284 - We'll look at the rightmost digits giving us 84, which is divisible by 4, so 1284 is also divisible by 4.258 - The two rightmost digits give us 58, which is not divisible by 4. Hence 258 is also not divisible by 4.
Any number is divisible by any other number. However, I am guessing you are actually asking if 416 is evenly divisible by 5, and that answer would be no. A quick way to tell if any natural number is evenly divisible by 5, is to see if the last (rightmost) digit is a 5 or a 0. If it is, the answer is yes. Otherwise, the answer is no.
Rightmost column is called the ones.
"rightmost " is not a proper word, however it's meaning is 'furthest to the right'
Since there are an infinite number of these, you cannot list them, but here's how you tell: Any non-zero integer which has one of the following in the ones place (rightmost digit): either a 5 or a 0
host portion.
22
You can always tell a number is divisible by eight by checking to see if the rightmost three digits are. If they are, then the whole number is. With that in mind then, you can make up all sorts of such numbers by just grabbing some combination that you know is divisible by eight, and tacking additional digits on to the left. For example, we know that 128 is divisible by eight (128 / 8 = 16), so any of these numbers are as well: 89469128 13948128 90173128 The number of digits doesn't matter, so we also know that a ridiculously large number like this one: 25098174509840582437575109283102477568758134024 is divisible by eight, because the last three digits are.
The leftmost point is the minimum value.The rightmost point is the maximum value.The difference between them is the range.The leftmost point is the minimum value.The rightmost point is the maximum value.The difference between them is the range.The leftmost point is the minimum value.The rightmost point is the maximum value.The difference between them is the range.The leftmost point is the minimum value.The rightmost point is the maximum value.The difference between them is the range.
In parsing algorithms, the difference between leftmost and rightmost derivation lies in the order in which non-terminal symbols are replaced with terminal or non-terminal symbols. In a leftmost derivation, the leftmost non-terminal symbol is always replaced first, while in a rightmost derivation, the rightmost non-terminal symbol is replaced first. This affects the structure and order of the resulting parse tree.
The rightmost derivation parse tree for the keyword "algorithm" is a tree structure that shows the order in which the grammar rules are applied to generate the keyword.
For 30 hosts, you need 5 bits (or more) for the host. That is, the 5 (or mor) rightmost bits in the subnet mask must be equal to zero; the remaining bits are 1, like this:11111111.11111111.11111111.11100000Now, just convert this to decimal (convert each of the parts, separated by dots, using Windows scientific calculator, or another scientific calculator).For 30 hosts, you need 5 bits (or more) for the host. That is, the 5 (or mor) rightmost bits in the subnet mask must be equal to zero; the remaining bits are 1, like this:11111111.11111111.11111111.11100000Now, just convert this to decimal (convert each of the parts, separated by dots, using Windows scientific calculator, or another scientific calculator).For 30 hosts, you need 5 bits (or more) for the host. That is, the 5 (or mor) rightmost bits in the subnet mask must be equal to zero; the remaining bits are 1, like this:11111111.11111111.11111111.11100000Now, just convert this to decimal (convert each of the parts, separated by dots, using Windows scientific calculator, or another scientific calculator).For 30 hosts, you need 5 bits (or more) for the host. That is, the 5 (or mor) rightmost bits in the subnet mask must be equal to zero; the remaining bits are 1, like this:11111111.11111111.11111111.11100000Now, just convert this to decimal (convert each of the parts, separated by dots, using Windows scientific calculator, or another scientific calculator).