School

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

Final Answer:

n =  301

Step-by-step explanation:

$$\begin{gathered} n<500 \\ (n-1)|3 \\ (n-1)|4 \\ (n-1)|5 \\ (n-1)|6 \\ n|7 \\ 3...\text{ prime number} \\ 4=2^2 \\ 5...\text{ prime number} \\ 6=2\cdot 3 \\ LCM(3,4,5,6)=2^2\cdot 3\cdot 5=60 \\ {n}_0=LCM(3,4,5,6)=60 \\ {n}_1=1\cdot n_0+1=1\cdot 60+1=61 \\ {t}_1=n_1/7=61/7=\frac{61}{7}=8\frac{5}{7}\doteq 8.7143 \\ {n}_2=2\cdot n_0+1=2\cdot 60+1=121 \\ {t}_2=n_2/7=121/7=\frac{121}{7}=17\frac{2}{7}\doteq 17.2857 \\ {n}_3=3\cdot n_0+1=3\cdot 60+1=181 \\ {t}_3=n_3/7=181/7=\frac{181}{7}=25\frac{6}{7}\doteq 25.8571 \\ {n}_4=4\cdot n_0+1=4\cdot 60+1=241 \\ {t}_4=n_4/7=241/7=\frac{241}{7}=34\frac{3}{7}\doteq 34.4286 \\ {n}_5=5\cdot n_0+1=5\cdot 60+1=301 \\ {t}_5=n_5/7=301/7=43 \\ {n}_6=6\cdot n_0+1=6\cdot 60+1=361 \\ {t}_6=n_6/7=361/7=\frac{361}{7}=51\frac{4}{7}\doteq 51.5714 \\ {n}_7=7\cdot n_0+1=7\cdot 60+1=421 \\ {t}_7=n_7/7=421/7=\frac{421}{7}=60\frac{1}{7}\doteq 60.1429 \\ {n}_8=8\cdot n_0+1=8\cdot 60+1=481 \\ {t}_8=n_8/7=481/7=\frac{481}{7}=68\frac{5}{7}\doteq 68.7143 \\ n=5\cdot n_0+1=5\cdot 60+1=301 \\ {d}_3=(n-1)/3=(301-1)/3=100 \\ {d}_4=(n-1)/4=(301-1)/4=75 \\ {d}_5=(n-1)/5=(301-1)/5=60 \\ {d}_6=(n-1)/6=(301-1)/6=50 \\ {d}_7=n/7=301/7=43 \end{gathered}$$

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