Yes, but only if you count a root at the tangent as a double root.
4
It is called the digital sum. If you keep doing it until you reach a single digit answer, it is called a digital root. The term digital root is sometimes also used for the digital sum. For example, 3042 gives the digital sum of 3+0+4+2 = 9 which is also its digital root. but 3047 gives the digital sum of 3+0+4+7 = 14 and the digital root is 1+4 = 5.
There is none. Any number formed from 3 identical digits must have a digital root of 0, 3 or 6. These are, therefore, the only possible remainders when divided by 9. Note: The digital root of a number is the sum of all its digits. If the answer is greater than 9, repeat the process.
A trick to see if a number is divisible by 3, 6, or 9 is to use something called the digital root.First, add all the digits up:3+8+5+2 = 18We want a single digit, so add the digits in 18 (from the previous step) up:1+8 = 9The nine (a single digit) is called the digital root.If a digital root of a number is equal to 3, 6, or 9 than the number is divisible by 3.If a digital root of a number is equal to 3, 6, or 9 AND the number is even, than the number is divisible by 6.If a digital root of a number is equal to 9, than the number is divisible by 9.The digital root of 3852 is 9, and 3852 is even, so using the rules above the number is divisible by 3, 6, AND 9.
There an infinity of such numbers. One test for divisibility by 9 depends on the digital root. The digital root of any number is the sum of all its digits. If the digital root of a number (or the digital root of its digital root) is divisible by 9 then the original number is, and if the DR is not then the original number is not.
Yes, but only if you count a root at the tangent as a double root.
4
Any number whose digital root is 9.To find the digital root of a number,add together all its digits.if the answer is bigger than 9, go back to step 1. Otherwisethe sum is the digital root.
When its digital root is 9 (divisible by 9) and it is even (divisible by 2). The digital root of a number is calculated by adding up all the digits in a number. If the answer is greater than 9, then repeat until you get an answer that is a single digit.
It is called the digital sum. If you keep doing it until you reach a single digit answer, it is called a digital root. The term digital root is sometimes also used for the digital sum. For example, 3042 gives the digital sum of 3+0+4+2 = 9 which is also its digital root. but 3047 gives the digital sum of 3+0+4+7 = 14 and the digital root is 1+4 = 5.
There is none. Any number formed from 3 identical digits must have a digital root of 0, 3 or 6. These are, therefore, the only possible remainders when divided by 9. Note: The digital root of a number is the sum of all its digits. If the answer is greater than 9, repeat the process.
A trick to see if a number is divisible by 3, 6, or 9 is to use something called the digital root.First, add all the digits up:3+8+5+2 = 18We want a single digit, so add the digits in 18 (from the previous step) up:1+8 = 9The nine (a single digit) is called the digital root.If a digital root of a number is equal to 3, 6, or 9 than the number is divisible by 3.If a digital root of a number is equal to 3, 6, or 9 AND the number is even, than the number is divisible by 6.If a digital root of a number is equal to 9, than the number is divisible by 9.The digital root of 3852 is 9, and 3852 is even, so using the rules above the number is divisible by 3, 6, AND 9.
Do you mean like a double floating point number, which is a complex number; or a double matrix type number like in the related link on springerlink.com
Subtract the remainder which is left when the digital root is divided by 9.
When you double a number, the square root of the new one is sqrt(2) = 1.4142 times the square root of the original one.
/* program to find Square Root of a given number. */ #include #include double Abs(double Nbr) { if( Nbr >= 0 ) return Nbr; else return -Nbr; } double SquareRoot(double Nbr) { double Number = Nbr / 2; const double Tolerance = 1.0e-7; do Number = (Number + Nbr / Number) / 2; while( Abs(Number * Number - Nbr) > Tolerance); return Number; } int main() { double Number ; double Nbr; clrscr(); printf("\n Enter any number:"); scanf("%lf",&Number); Nbr=SquareRoot(Number); printf("The square root of %.2lf is %.2lf", Number,Nbr); getch(); }