Possibilities 5822

Peter forgot the four-digit code to his school locker lock. Fortunately, his mother remembered some information about him. He knows that the first binary number is divisible by 15 and the second by 7. However, Peter is a big loser, so he has to try all the possibilities (including the possibility of 0000). How many attempts did Peter take to open the lock?

Final Answer:

n =  105

Step-by-step explanation:

$$\begin{gathered} n_1=0000 \\ {n}_2=0007 \\ {n}_3=0014 \\ {n}_4=0021 \\ {n}_5=0028 \\ {n}_6=0035 \\ {n}_7=0042 \\ {n}_8=0049 \\ {n}_9=0056 \\ {n}_{10}=0063 \\ {n}_{11}=0070 \\ {n}_{12}=0077 \\ {n}_{13}=0084 \\ {n}_{14}=0091 \\ {n}_{15}=0098 \\ {n}_{16}=1500 \\ {n}_{17}=1507 \\ {n}_{18}=1514 \\ {n}_{19}=1521 \\ {n}_{20}=1528 \\ {n}_{21}=1535 \\ {n}_{22}=1542 \\ {n}_{23}=1549 \\ {n}_{24}=1556 \\ {n}_{25}=1563 \\ {n}_{26}=1570 \\ {n}_{27}=1577 \\ {n}_{28}=1584 \\ {n}_{29}=1591 \\ {n}_{30}=1598 \\ {n}_{31}=3000 \\ {n}_{32}=3007 \\ {n}_{33}=3014 \\ {n}_{34}=3021 \\ {n}_{35}=3028 \\ {n}_{36}=3035 \\ {n}_{37}=3042 \\ {n}_{38}=3049 \\ {n}_{39}=3056 \\ {n}_{40}=3063 \\ {n}_{41}=3070 \\ {n}_{42}=3077 \\ {n}_{43}=3084 \\ {n}_{44}=3091 \\ {n}_{45}=3098 \\ {n}_{46}=4500 \\ {n}_{47}=4507 \\ {n}_{48}=4514 \\ {n}_{49}=4521 \\ {n}_{50}=4528 \\ {n}_{51}=4535 \\ {n}_{52}=4542 \\ {n}_{53}=4549 \\ {n}_{54}=4556 \\ {n}_{55}=4563 \\ {n}_{56}=4570 \\ {n}_{57}=4577 \\ {n}_{58}=4584 \\ {n}_{59}=4591 \\ {n}_{60}=4598 \\ {n}_{61}=6000 \\ {n}_{62}=6007 \\ {n}_{63}=6014 \\ {n}_{64}=6021 \\ {n}_{65}=6028 \\ {n}_{66}=6035 \\ {n}_{67}=6042 \\ {n}_{68}=6049 \\ {n}_{69}=6056 \\ {n}_{70}=6063 \\ {n}_{71}=6070 \\ {n}_{72}=6077 \\ {n}_{73}=6084 \\ {n}_{74}=6091 \\ {n}_{75}=6098 \\ {n}_{76}=7500 \\ {n}_{77}=7507 \\ {n}_{78}=7514 \\ {n}_{79}=7521 \\ {n}_{80}=7528 \\ {n}_{81}=7535 \\ {n}_{82}=7542 \\ {n}_{83}=7549 \\ {n}_{84}=7556 \\ {n}_{85}=7563 \\ {n}_{86}=7570 \\ {n}_{87}=7577 \\ {n}_{88}=7584 \\ {n}_{89}=7591 \\ {n}_{90}=7598 \\ {n}_{91}=9000 \\ {n}_{92}=9007 \\ {n}_{93}=9014 \\ {n}_{94}=9021 \\ {n}_{95}=9028 \\ {n}_{96}=9035 \\ {n}_{97}=9042 \\ {n}_{98}=9049 \\ {n}_{99}=9056 \\ {n}_{100}=9063 \\ {n}_{101}=9070 \\ {n}_{102}=9077 \\ {n}_{103}=9084 \\ {n}_{104}=9091 \\ {n}_{105}=9098 \\ n=15\cdot 7=105 \end{gathered}$$

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