num=32767
MsgBox(len(num))
<html> <script language="vbscript"> n=cint(inputbox("Enter a number")) dim f f=1 if n<0 then Msgbox "Invalid number" elseif n=0 or n=1 then MsgBox "The factorial of given number "&n&" is :"&f else for i=n to 2 step -1 f=f*i next MsgBox "The factorial of given number "&n&" is :"&f end if </script> </html>
I assume you mean is an if the number is an integer multiple of 3i am unfamiliar with C but the theory would be,find if a is integer multiple of 3b=a/3b==round(b).if 1 "yes"else "no"this is an inefficiency way but will get the job done
It doesn't have length, but you can use sizeof to find out its size.
: Find the percentage error if 625.483 is approximated to 3 significant digits?
The tricky part is getting the individual digits. There are basically two ways to do this: 1) Convert the number to a string, and use string manipulation to get the individual digits. 2) Repeatedly divide the number by 10. The digit is the remainder (use the "%" operator). To actually get the highest digit, initially assume that the highest digit is zero (store this to a variable, called "maxDigit" or something similar). If you find a higher digit, replace maxDigit by that.
Any number using each of the digits once will be a multiple of 3: eg 1597864302
1/9
enter the number whose digits are to be added num is the given value num=0! k=num%10 sum=sum=k k=num/10 num=k print the sum of the digits
To find an integer you ask your teacher.
You can try to subtract the first few digits that are displayed from the answer, but note that calculators work to a limited precision so you may not be able to get much more than 10 to 12 digits this way and if the decimal has more than 12 digits you will not be able to find all the digits. The only solution is to do the long division (by hand, using the digits displayed by the calculator, but when you run out of digits in the calculator, you can start a new division using the remainder so far to continue for the next few digits). Alternatively, find a calculator which works to more precision.
Yes. 117/9=13. An easy test for divisibility by nine (for an integer of any length) is to add all of the digits. If the sum is nine or a number divisible by nine, then the integer is divisible by nine. In this case, 1+1+7=9, so 117 is divisible by nine. (Be careful, this test fo divisibility only works generally for divisibility by three -- i.e., an integer is divisible by three if and only if the sum of its digits equals three or a multiiple of three-- and for nine.) To find if an integer is divisible by 4, you can check if the last two digits are. If they are, it is. To check if an integer is divisible by 6, you must make sure that the integer is divisible by both 3 and 2. If you want to check if an integer is divisible by 2, just make sure it's even. Any integer divisible by 10 will end in zero. Any integer divisible by 5 will end in either 5 or 0. This is an important part of pre-algebra, as well as algebra.
The absolute value of a number is expressed with the symbol |. To show you want to find the absolute value of an integer(using x as the integer) you would do this |x|. Examples- |2| = 2 |-9| = 9 |325| = 325 |-457245| = 457245
71 is, itself an integer but to find a difference you require 3 numbers.71 is, itself an integer but to find a difference you require 3 numbers.71 is, itself an integer but to find a difference you require 3 numbers.71 is, itself an integer but to find a difference you require 3 numbers.
Well, 47 49 51 53 are four consecutive odd numbers those total squared has for identical digits. 40000.... The square root of any number that is only four digits long all containing the same digit has a value that is not an integer.
Assuming the answer is a positive integer then the range of integers that round to 5600 to the nearest hundred is 5550 to 5649. As the first two digits sum to 10 or 11, then the questioner may be able to find the answer by elimination.
2
find the sum of 2 and 2