[Guest]

The square of each of the base ten positive integers from 1 to 2010 inclusively are written successively without commas or spaces, resulting in the following pattern:

14916253649............4040100

Determine the total number of times that the digit 9 appear in the abovementioned pattern.

Edited January 27, 2010 by K Sengupta

[Guest]

1178

[Guest]

function getIntegerCountFromSquaredSequence(num, range_min, range_max) {

var n_count = 0;

for(var i=range_min;i <= range_max;i++) {

var s_temp = (i * i).toString();

for(var j=0;j<s_temp.length;j++) {

if (s_temp.charAt(j)==num.toString()) {

n_count++;

}

}

}

return n_count;

}

alert(getIntegerCountFromSquaredSequence(9, 1, 2010));

The previous poster was correct.

[Guest]

Using Excel and brute force calculation - 1178 times.

[unreality]

with java

public static int numNines(int cap)

    {

        int num=0;

        for (int n=1; n<=cap; n++) num += ninesIn(n*n);

        return num;

    }

    public static int ninesIn(int n)

    {

        int nums=0;

        while (n > 0) {

            if (n % 10 == 9) nums++;

            n /= 10;

        }

        return nums;

    }

pass 2010 to "numNines" ... answer 1178

[Guest]

Very neat, but I'm waiting for the intended mathematical, rather than programmatic, solution from KS.

My weak number theory knowledge expired 20yrs ago, so I'm not even going to attempt it!