Remainder

What is the remainder of the division of natural numbers 293 and 7?

Correct result:

z =  6

Solution:

p=293/7=2937≐41.8571 p1=⌊p⌋=⌊41.8571⌋=41  z=293−p1⋅ 7=293−41⋅ 7=6p=293/7=\dfrac{ 293 }{ 7 } \doteq 41.8571 \ \\ p_{1}=\lfloor p \rfloor=\lfloor 41.8571 \rfloor=41 \ \\ \ \\ z=293 - p_{1} \cdot \ 7=293 - 41 \cdot \ 7=6

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