if the square of a number ends with 10 zeroes, how many zeroes will the number have?
we know that if a number ends with n zeros then its square will have 2n zeroes conversely, if square of a number have 2n zeros at their ends then the number will have n zeroesthe square of a number ends 10 zeroes, then the number will have `(10)/(2)` = 5 zeroes
zero express as a sum of perfect square 16 a square + b square + 9 c square minus 4 a b minus 3 b c - 12 ca? * If the number ends with a $0$ then there are $9$ choices for the first digit, $8$ for the second and $7$ for the third, which makes $1\times9\times8\times7= 504$ possibilities.
* If the number is even ending with something else than $0$ then there are $4$ choices for the last digit, $8$ choices for the first digit (no $0$ nor the last digit), $8$ for the second digit and $7$ for the third digit, which makes $4\times8\times8\times7 =1792$


Together, this gives $2296$ numbers with $4$ distinct digits that are even. Note that this does not allow leading $0$, as you see to want it based from the question.


Correct Answer: $B$ 2, 0 ( any two integers with one of them as 0). 1, 2, 7, 8, 10 so finding the square root by vedic method, four digits are in the number, so the digits in it’s square root will be 2. the quotient is 46, so the square root is 46.hint –i. four digits in the number so two digits in the square root.ii. first square root digit = 4iii. 21 - = 5 = remainderiv. pair of digits 16 was descended before 5. so new dividend = 516.v. divisor = 4 2 = 8, i.e., double of square root digit.vi. 86 divides 516 by 6, so write 6 ahead digit 4 in the quotient.vii. 516 – 86 6 = 516 – 516 = 0remainder = 0, square root = quotient = 46. You need to give the user the option of typing in a value that represents that there are no more values to enter. Perhaps give a prompt like:



```
Please enter a value.  Enter -999999 if you have no more values [B]:[/B]
```

As for the ten (or fewer) values, you should probably have an array and a loop instead of ten separate variable names and ten separate cin statements. So you're looking at something like this:



```
double dValue[10];
    int numEntries = 0;
    double sum = 0;
    
    while (numEntries < 10)
    {
          cout << "Enter number (-999999 to stop) : ";
          cin >> dValue[numEntries];
          // check whether dValue[numEntries] == -999999
          // if so, short-circuit the loop with the "break" command
            
          // increment numEntries
          // adjust sum
    }
```

As a side note, I picked -999999 as the value to represent no more values because



1. It's an integer, so it can be represented exactly as a double (no round-off error using == to compare).
2. It's a number that, chances are, people wouldn't actually want to add, so it's sort of "out of the way". Chances are, few people want to add -999999. If they do, they can't use this program! The additive inverse of 0 is 0. equating the expression with 0,3x2 + 5x + 2 = 0on factorising it further,3x2 + 3x + 2x + 2 = 03x(x + 1) + 2(x + 1) = 0(3x + 2) (x + 1) = 0 x = - or x = -1 the zeroes of 3x2 + 5x + 2 are – and -1. answer: 0