#2**+1 **

Challenge accepted

Let the girls be a,b,c,d, and e where a is the lightest and e is the heaviest.

If I add all those paired weithgts together I get 1212 pounds

Each girl is included 4 times so

4(a+b+c+d+e) = 1212 pounds

a+b+c+d+e = 303

The average weigt of the children is 60.6 pounds

The average weight of each pair of girls is 121.2 pounds

So now it is important to know how far each pair of girls **deviate from the average**

score | score -121.2 | ||

129 | D+E | 7.8 | |

125 | C+E | 3.8 | D=C+4 |

124 | ? | 2.8 | |

123 | ? | 1.8 | |

122 | ? | 0.8 | |

121 | ? | -0.2 | |

120 | ? | -1.2 | |

118 | ? | -3.2 | |

116 | A+C | -5.2 | C=B+2 |

114 | A+B | -7.2 | |

sum | 0 |

A | B | C | D | E |

A | B | B+2 | C+4=B+6 | E |

A+B = -7.2 So A= -7.2 -B

D+E= 7.8 so B+6+E = 7.8 so E = 1.8 - B

A | B | C | D | E |

-7.2 - B | B | B+2 | C+4=B+6 | 1.8 - B |

These are all deviations so they have to add up to 0.

-7.2 - B +B +B+2 +B+6+1.8-B=0

2.6+B=0

B=-2.6

A | B | C | D | E | |

distance from mean | -7.2 - -2.6 = -4.4 | -2.6 | -2.6+2=-0.6 | -2.6+6 =3.4 | 1.8 - -2.6= 4.4 |

Girls weight | 60.6-4.4=56 | 60.6-2.6=58 | 60.6-0.6=60 | 60.6+3.4=64 | 60.6+4.4=65 |

**So the girls weigh 56, 58, 60, 64 and 65 pounds respectively. **

Is that how you did it Chris?

Melody
4 hours ago

Sep 23, 2017

#1**0 **

\(\sqrt{25}=5\)

Since we only need to calculate the integers, the approximation for \(\sqrt{140}\) does not need to be too accurate. I know that \(12^2=144\) and \(11^2=121\), so \(\sqrt{140}\) can be rounded down to 11.

There is a formula that allows one to calculate the running total of consecutive integers from one to the last integer. It is \(\frac{n(n+1)}{2}\), where *n* is the largest number in the set, in this case 11. However, we have a bit of an issue.

This formula only works when the consecutive integers start at 1. In this case, the integers start at 5. Therefore, we must subtract the sum of the integers from 1 to 4. We can use the same formula of \(\frac{n(n+1)}{2}\), where n=4.

\(\frac{11(11+1)}{2}-\frac{4(4+1)}{2}\)

Find the result of this.

\(\frac{11(11+1)}{2}-\frac{4(4+1)}{2}\) | Do what is inside the parentheses first. |

\(\frac{11*12}{2}-\frac{4*5}{2}\) | |

\(11*6-2*5\) | |

\(66-10\) | |

\(56\) | |

TheXSquaredFactor
9 hours ago

#1**+2 **

The only answer is 5. I did this by examining a table of values.

There is bound to be a more elegant way of doing it but I do not know what it is.

\(\left\lfloor n^2/4 \right\rfloor - \lfloor n/2 \rfloor^2 = 2\)

n | -10 | -9 | -8 | -7 | -6 | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

\(\left\lfloor n^2/4 \right\rfloor \) | 25 | 20 | 16 | 12 | 9 | 6 | 4 | 2 | 1 | 0 | 0 | 0 | 1 | 2 | 4 | 6 | 9 | 12 | 16 | 20 | 25 |

\(\lfloor n/2 \rfloor^2\) | 25 | 25 | 16 | 16 | 9 | 9 | 4 | 4 | 1 | 1 | 0 | 0 | 1 | 1 | 4 | 4 | 9 | 9 | 16 | 16 | 25 |

\(\left\lfloor n^2/4 \right\rfloor - \lfloor n/2 \rfloor^2\) | 0 | -5 | 0 | -4 | 0 | -3 | 0 | -2 | 0 | -1 | 0 | 0 | 0 | 1 | 0 | 2 | 0 | 3 | 0 | 4 | 0 |

So I can see that for all even numbers, positive and negative the answer will be zero.

For the odd numbers a pattern emerges.

It is clear to me that the only integer value of n to make this true is 5

Melody
Sep 23, 2017 7:20:23 AM

Sep 22, 2017