Chmod Math
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From Julio Sepia:
Write the number 755 as the sum of five consecutive odd numbers.
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11 Answers Challenge Your Friends
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From Julio Sepia:
Write the number 755 as the sum of five consecutive odd numbers.
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This is the type of stuff that Linq excels in.
Quick and dirty version:
var eval = from i in Enumerable.Range(0, 1000)
where i % 2 != 0
let cs = Enumerable.Range(i, 5 * 2).Where(j => j % 2 != 0)
let sum = cs.Sum()
select new { Sum = sum, Numbers = cs};
var result = eval.First(i => i.Sum == 755);
Console.WriteLine(string.Join(” + “, result.Numbers) + ” = ” + result.Sum);
Prints:
147 + 149 + 151 + 153 + 155 = 755
(x)+(x+2)+(x+4)+(x+6)+(x+8)=755
5*x+20=755
5*x=735
x=147
755 = 147 + 149 + 151 + 153 + 155
Less Stupid version:
static IEnumerable ConsecutiveOdds(int target, int noConsecutives)
{
var midPoint = target / (double)noConsecutives;
var start = (int)(midPoint – Math.Floor(noConsecutives / 2.0) * 2);
return Enumerable.Range(start, noConsecutives * 2).Where(i => i % 2 == 0);
}
>>> 755/5
151
>>> 147+149+151+153+155
number = 755
number_of_odds = 5
mid = number / number_of_odds
odd_numbers = [x for x in xrange(mid-4,mid+5) if x % 2 != 0]
mid_index = odd_numbers.index(mid)
print “%d + %d + %d + %d + %d = %d” % \
(odd_numbers[mid_index-2],odd_numbers[mid_index-1],odd_numbers[mid_index],
odd_numbers[mid_index+1],odd_numbers[mid_index+2],number)
prints:
147 + 149 + 151 + 153 + 155 = 755
Came to post R G’s solution.
idiomatic java version :
——————————-
import junit.framework.TestCase;
public class ChmodMath extends TestCase{
public int[] consecutiveOdd(int target,int numberOfConsecutives){
for(int i = 0; i < target; i++){
if(i % 2 != 0){
int sum = 0;
int[] consecutiveOff = new int[numberOfConsecutives];
for(int j=i, index=0;j< i+(numberOfConsecutives*2); j+=2, index++){
consecutiveOff[index] = j;
sum += j;
}
if(sum == target){
return consecutiveOff;
}
}
}
return null;
}
public void testConsecutiveOdd() throws Exception {
int[] odds = consecutiveOdd(755, 5);
int total = 0;
for(int odd : odds){
System.out.println(odd);
total += odd;
}
assertEquals(755, total);
}
}
————————
Agree with R G.
Yup, also came to post R.G.’s solution.
It¡¦s in point of fact a nice and useful piece of info. I am glad that you shared this helpful information with us. Please keep us up to date like this. Thank you for sharing.
public class puzzle755 {
public int siguienteNumeroImpar(int actual){
int respuesta=0;
if ((actual % 2) == 0){
actual++;
}
System.out.println(“actual : ” + actual);
return actual;
}
public boolean calculando(){
int suma = 0,digito1,digito2,digito3,digito4,digito5 =0;
for (int i=0; suma <= 756; i++){
digito1= this.siguienteNumeroImpar(i +1);
digito2 =this.siguienteNumeroImpar(digito1+1);
digito3 =this.siguienteNumeroImpar(digito2+1);
digito4 =this.siguienteNumeroImpar(digito3+1);
digito5 =this.siguienteNumeroImpar(digito4+1);
suma =digito1 + digito2 +digito3+digito4+digito5;
System.out.println("suma : " +suma);
if (suma== 755){
System.out.println("r : " + digito1);
System.out.println("r : " + digito2);
System.out.println("r : " + digito3);
System.out.println("r : " + digito4);
System.out.println("r : " + digito5);
}
i ++;
}
return true;
}
public static void main(String args []){
System.out.println(new puzzle755().calculando());
}
}