271 can be written as $271=256+8+4+2+1=100001111_2$ It has a beauty of 5.

So we want the largest number less than 1000 with a beauty of 5. In other words we need the largest possible number less than 1000 which can be written as the sum of 5 different powers of 2

The largest power of 2 it can contain is 512, then 256, then 128 and so on...

So the number can be $n=512+256+128+64+32=992$ We are lucky that the first number we guessed is less than 1000. That is the answer. Or else we would have to consider smaller numbers by adding 16 instead of 32, and so on...

We can use the convenient Python method bin which converts an integer to a binary string in combination with Python's string method count to accomplish this task. First, we define a function beauty(x) like this:

def beauty(x):
  return bin(x).count("1")

Then, we evaluate beauty(271) and we get $5$ , therefore we can write a loop that solves the problem:

num = 0
for x in range(100, 1000):
  if x > num and beauty(x) == 5:
    num = x

If we run this loop, we find that num is $992$ .

Python Solution:

def digitSum(s):
    return sum([int(i) for i in s[2:]])

for i in range(1000)[::-1]:
    if digitSum(bin(271)) == digitSum(bin(i)):
        print i
        break

Python 3.4.1

one_count = lambda number: bin(number).count('1')

target = one_count(271)

print(max([n if (one_count(n) == target) else 0 for n in range(271, 1000)]))

Solution in C++ :

    unsigned int binOnes(unsigned int n)
    {
        if(n < 2)
            return n;
        return (n%2) + binOnes(n/2);
    }
    int main()
    {
        int target = binOnes(271) , num;
        for(num = 999 ; binOnes(num) != target ; --num);
        cout << num << endl;
        return 0;
    }

No. of ones in 271 (0000000 1 0000 1111) = 5 The first largest number with 5 ones till 9th bit can be

0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 =496

The above thing is checked because in this 9th bit is acting as MSB . Moving these ones to 10th bit i.e the next MSB like:

0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0  = 992

this is the largest number

for(i=0; cpyDecNumber>0; i++) { quotient = cpyDecNumber % 2; binaryArray1[i] = quotient; cpyDecNumber = cpyDecNumber / 2; } arrayIndex = i-1; j = arrayIndex; printf("\nBinary Equivalent of the given number is :\n"); while(j>=0) { if(binaryArray1[j] == 1) { beautyCounter++; } printf("%d", binaryArray1[j]); j = j-1; } beautyGiven = beautyCounter; printf("\nHence the Beauty of the given number is : %d\n", beautyGiven); printf("\nWe are processing for finding the largest 3-digit number with same beauty!\n");

for (j=8; j>(8-beautyGiven); j--)
{
    binaryArray2[j] = 1;
}
for(i=8; i>0; i--)
{
    printf("%d", binaryArray2[i]);
    sameBeauty = sameBeauty + ((pow(2,i+1))*binaryArray2[i]);
}
printf("\n%d", sameBeauty);

I have written the code in python:

Solution:

for i in xrange(999,99,-1):

    if bin(i)[2:].count('1') == 5:

        print i

        break

The answer is:

992

def beautiful(n):

binary=str(bin(n))[2:]

sum=0

for i in range(0,len(binary)):

    if int(binary[i])==1:

        sum = sum + int(binary[i])    

return sum

def smallest(digits):

if digits==1:

    return 1

return pow(10,digits-1)

def largest(digits):

return pow(10,digits)-1

def find beautiful as(asNumber,noOfDigits):

asNum=asNumber

for i in range(smallest(noOfDigits),largest(noOfDigits)):

    b = beautiful(i)

    if b==beautiful(asNum):

        asNum=i

return asNum

print find beautiful as(271,3)

include <stdio.h>

int main() { int i, j, a=0; for (i=999;i>99;i--) { for (j=i;j>0;j=j/2) { if (j%2==1) a++; } if (a==5) { printf("%d", i); break; } a=0;

} }

answer = 992

#    x = 999
#    sumX = 0
#    while x > 271:
#        binaryX = bin(x)
#        for s in binaryX:
#            if s == '1':
#                sumX += 1
#        if ( sumX == 5):
#            print x
#            break
#        x = x - 1
#        sumX = 0

This solution is purely mathematics - no programming.

$271_{10}=100001111_2$

which has 5 '1's.

We need to find the largest three-digit number with five ones. Note that we cannot go over 512, since 1024 is the next highest power of 2 and that is over 1000.

The largest we can go is therefore

$512+256+128+64+32=992$

Since we cannot go any larger and this number satisfies our conditions, our answer is $\boxed{992} \ \blacksquare$

for the first write $271$ in binary representation that is $271=100001111$ , so $271$ beautiful as $5$ Just find the largest $n$ such that $2^n<1000$ that is $n=9$ and then find the largest $n$ such that $2^n<1000-2^9$ that is $n=8$ and then find the largest $n$ such that $2^n<1000-2^9-2^8$ that is $n=7$ and then find the largest $n$ such that $2^n<1000-2^9-2^8-2^7$ that is $n=6$ and the fifth digit is find the largest $n$ such that $2^n<1000-2^9-2^8-2^7-2^6$ we can use this method because $1+2+2^2 +...+2^{n-1}=2^n-1<2^n$ .

solution

def to_bin(n): #converting a number to binary
    k=10     
    p=0
    while k>=0:
        if n/(2**k) >= 1:
            n=n-(2**k)
            p=p*10+1
        else:
            p=p*10
        k=k-1
    return p

def count_beauty(n): # counting beauty of n using its binary
    binofn=to_bin(n)
    beauty=0
    for ele in str(binofn):
            if ele=='1':
            beauty+=1

    return beauty

b=count_beauty(271)     # ''271''
number=1000             # counting backwards from 1000

while number>=0:
    if count_beauty(number)==b:
        print  number
        break
    number=number-1

I simply used brute force to get the answer. My python code.

R = 271
target_beauty = bin(R).count('1')
for i in range(R,1000):
    beauty = bin(i).count('1')
    if beauty == target_beauty:
         if i > R:
            R = i
print(R)

271 in the binary representation is 100001111. Sum of number of '1's is 5. So the largest 3-digit number must have sum of number of '1's is 5. The number 111110000 is 9 digits <=> In the decimal representation is 496. We try add 1 number 0 to the last, we have 1111100000 is 10 digits <=> In the decimal representation is 992. So the largest 3-digit number is 992

The interesting thing is to convert from decimal to binary, here is my C++ Code:

            void binary(int a)
            {
                vector <int> binary_vector;
                long binary_number, dividend;

                cout << "\nDecimal: " << a;
                dividend = a;

                while (dividend >= 1)
                {
                    binary_number = dividend % 2;
                    dividend /= 2; 
                    binary_vector.push_back(binary_number);
                }

                cout << " Binary: ";
                for (int i = binary_vector.size() - 1; i >= 0; i--)
                    cout << binary_vector[i];
            }

            int _tmain(int argc, _TCHAR* argv[])
            {
                int h, j;
                cin >> h;
                cin >> j;

                for (h; h > j; h--)
                {
                    binary(h);
                }

                _getch();

                return 0;
            }

//write include iostream statement, then write include string statement, then copy the code and paste

using namespace std;

typedef unsigned short int USHORT;

int toPower (const int number, const int exponent) {

if (number == 1 || exponent == 0) { return (1); }

else if (number == 0) { return (0); }

else if (exponent == 1) { return (number); }

else {

int buffer (number);

for (int i = 0; i < (exponent - 1); i++) {

  buffer *= number;

}

return (buffer);

}

}

int getMaxExponent (const int& number, const int base) {

int tester (0), counter (0);

while (tester <= number) {

++counter;

tester = (toPower (base, counter));

}

--counter;

return (counter);

}

string convertToBin (int usrInput) {

string subject;

int buffer (0), resultLength (0), counter (0);

int maxExponent = getMaxExponent (usrInput, 2);

while (maxExponent > -1) {

buffer = toPower (2, maxExponent);

if (usrInput >= buffer) {

  usrInput -= buffer;

  subject += "1";

}

else {

  subject += "0";

}

--maxExponent;

}

return (subject);

}

int main (int argc, char* argv []) {

const int numBeauty (5);

int counter (0), greatest (0), resultLength (0);

string result;

for (int i = 100; i <= 999; i++) {

if (i == 271) { continue; }

else {

  result = convertToBin (i);

  for (int j = 0; j < (result.length ()); j++) {

if (result [j] == '1') { counter++; }

  }

  if (counter == numBeauty && (i > greatest)) {

greatest = i;

  }

  counter = 0;

}

}

cout << greatest << endl;

return (0);

}