# School

Less than 500 pupils attend school. When it is sorted into pairs, one pupil remains. Similarly, when sorted into 3, 4, 5 and 6 members team one remains. Sorted to seven members teams, no left behind. How many pupils are attending this school?

Result

n =  301

#### Solution:

$n<500 \ \\ (n-1)%3 \ \\ (n-1)%4 \ \\ (n-1)%5 \ \\ (n-1)%6 \ \\ n%7 \ \\ 3 = 3 \\ 4 = 2^2 \\ 5 = 5 \\ 6 = 2 \cdot 3 \\ LCM(3, 4, 5, 6) = 2^2 \cdot 3 \cdot 5 = 60\\ \ \\ \ \\ n_{ 0 } = LCM(3,4,5,6) = 60 \ \\ \ \\ n_{ 1 } = 1 \cdot \ n_{ 0 }+1 = 1 \cdot \ 60+1 = 61 \ \\ t_{ 1 } = n_{ 1 } / 7 = 61 / 7 = \dfrac{ 61 }{ 7 } \doteq 8.7143 \ \\ \ \\ n_{ 2 } = 2 \cdot \ n_{ 0 }+1 = 2 \cdot \ 60+1 = 121 \ \\ t_{ 2 } = n_{ 2 } / 7 = 121 / 7 = \dfrac{ 121 }{ 7 } \doteq 17.2857 \ \\ \ \\ n_{ 3 } = 3 \cdot \ n_{ 0 }+1 = 3 \cdot \ 60+1 = 181 \ \\ t_{ 3 } = n_{ 3 } / 7 = 181 / 7 = \dfrac{ 181 }{ 7 } \doteq 25.8571 \ \\ \ \\ n_{ 4 } = 4 \cdot \ n_{ 0 }+1 = 4 \cdot \ 60+1 = 241 \ \\ t_{ 4 } = n_{ 4 } / 7 = 241 / 7 = \dfrac{ 241 }{ 7 } \doteq 34.4286 \ \\ \ \\ n_{ 5 } = 5 \cdot \ n_{ 0 }+1 = 5 \cdot \ 60+1 = 301 \ \\ t_{ 5 } = n_{ 5 } / 7 = 301 / 7 = 43 \ \\ \ \\ n_{ 6 } = 6 \cdot \ n_{ 0 }+1 = 6 \cdot \ 60+1 = 361 \ \\ t_{ 6 } = n_{ 6 } / 7 = 361 / 7 = \dfrac{ 361 }{ 7 } \doteq 51.5714 \ \\ \ \\ n_{ 7 } = 7 \cdot \ n_{ 0 }+1 = 7 \cdot \ 60+1 = 421 \ \\ t_{ 7 } = n_{ 7 } / 7 = 421 / 7 = \dfrac{ 421 }{ 7 } \doteq 60.1429 \ \\ \ \\ n_{ 8 } = 8 \cdot \ n_{ 0 }+1 = 8 \cdot \ 60+1 = 481 \ \\ t_{ 8 } = n_{ 8 } / 7 = 481 / 7 = \dfrac{ 481 }{ 7 } \doteq 68.7143 \ \\ \ \\ n = 5 \cdot \ n_{ 0 }+1 = 5 \cdot \ 60+1 = 301 \ \\ \ \\ d_{ 3 } = (n-1)/3 = (301-1)/3 = 100 \ \\ d_{ 4 } = (n-1)/4 = (301-1)/4 = 75 \ \\ d_{ 5 } = (n-1)/5 = (301-1)/5 = 60 \ \\ d_{ 6 } = (n-1)/6 = (301-1)/6 = 50 \ \\ d_{ 7 } = n/7 = 301/7 = 43$

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