Question

How many 4 digit numbers are divisible by 8 but not by 7?

Answer 1

$If\ a\ number\ is\ divisible\ by\ both\ 7\ and\ 8,\ that\ number\ will\ be\ a \\ multiple\ of\ 7 \times 8 = 56. \\ \\ First\ let's\ find\ no.\ of\ 4-digit\ numbers\ which\ are\ divisible\ by\ 8. \\ \\ Smallest\ 4-digit\ number\ is\ 1000. \\ \\ 1000\ divided\ by\ 8\ gives\ remainder\ 0. \\ \\ \therefore\ 1000\ is\ the\ smallest\ 4-digit\ number\ which\ is\ divisible\ by\ 8.$

$\\ \\ \\ Largest\ 4-digit\ number\ is\ 9999. \\ \\ 9999\ divided\ by\ 8\ gives\ remainder\ 7. \\ \\ \therefore\ 9999 - 7 = 9992\ is\ the\ smallest\ 4-digit\ number\ which\ is \\ divisible\ by\ 8. \\ \\ \\ Consider\ the\ 4-digit\ multiples\ of\ 8\ are\ in\ AP.$

$\\ \\ T_1 = 1000 \\ \\ T_n = 9992 \\ \\ d = 8 \\ \\ n = \frac{T_n - T_1}{d} + 1 \\ \\ = \frac{9992 - 1000}{8} + 1 = \frac{8992}{8} + 1 = 1124 + 1 = 1125 \\ \\ \\ \therefore\ There\ are\ 1125\ \ 4-digit\ multiples\ of\ 8.$

$\\ \\ \\ Let's\ find\ the\ no.\ of\ 4-digit\ numbers\ which\ are\ divisible \\ by\ 56. \\ \\ 1000\ divided\ by\ 56\ gives\ remainder\ 48. \\ \\ 56 - 48 = 8 \\ \\ \therefore\ 1000 + 8 = 1008\ is\ the\ smallest\ 4-digit\ number\ which\ is \\ divisible\ by\ 56. \\ \\ 9999\ divided\ by\ 56\ gives\ remainder\ 31. \\ \\ \therefore\ 9999 - 31 = 9968\ is\ the\ largest\ 4-digit\ number\ which\ is \\ divisible\ by\ 56.$

$\\ \\ \\ Consider\ the\ 4-digit\ multiples\ of\ 56\ are\ in\ AP. \\ \\ T_1 = 1008 \\ \\ T_n = 9968 \\ \\ d = 56 \\ \\ n = \frac{T_n - T_1}{d} + 1 \\ \\ = \frac{9968 - 1008}{56} + 1 = \frac{8960}{56} + 1 = 160 + 1 = 161 \\ \\ \\ \therefore\ There\ are\ 161\ \ 4-digit\ multiples\ of\ 56.$

$\\ \\ \\ Now\ subtract\ the\ no.\ of\ 4-digit\ multiples\ of\ 56\ from\ the\ no. \\ of\ 4-digit\ multiples\ of\ 8\ to\ get\ the\ answer. \\ \\ 1125 - 161 = \bold{964} \\ \\ \\ \therefore\ There\ are\ \bold{964}\ \ 4-digit\ numbers\ which\ are\ divisible\ by\ 8\ but \\ not\ by\ 7.$

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