Question Number 220858 by Rojarani last updated on 20/May/25
Commented by mr W last updated on 21/May/25
$$\mathrm{65}? \\ $$
Commented by Rojarani last updated on 21/May/25
$$\:{Sir},\:{I}\:{don}'{t}\:{know}\:{the}\:{answer}. \\ $$
Answered by mr W last updated on 21/May/25
$${number}\:{of}\:{numbers}\:{from}\:\mathrm{1}\:{to}\:{N}, \\ $$$${which}\:{are}\: \\ $$$${divisible}\:{by}\:\mathrm{3}\:{is}\:\lfloor\frac{{N}}{\mathrm{3}}\rfloor \\ $$$${divisible}\:{by}\:\mathrm{5}\:{is}\:\lfloor\frac{{N}}{\mathrm{5}}\rfloor \\ $$$${divisible}\:{by}\:\mathrm{7}\:{is}\:\lfloor\frac{{N}}{\mathrm{7}}\rfloor \\ $$$${divisible}\:{by}\:{both}\:\mathrm{5}\:{and}\:\mathrm{7}\:{is}\:\lfloor\frac{{N}}{\mathrm{35}}\rfloor \\ $$$${divisible}\:{by}\:\mathrm{5}\:{or}\:\mathrm{7}\:\left({or}\:{both}\right)\:{is}\:\lfloor\frac{{N}}{\mathrm{5}}\rfloor+\lfloor\frac{{N}}{\mathrm{7}}\rfloor−\lfloor\frac{{N}}{\mathrm{35}}\rfloor \\ $$$${we}\:{are}\:{given}: \\ $$$$\lfloor\frac{{N}}{\mathrm{3}}\rfloor=\lfloor\frac{{N}}{\mathrm{5}}\rfloor+\lfloor\frac{{N}}{\mathrm{7}}\rfloor−\lfloor\frac{{N}}{\mathrm{35}}\rfloor \\ $$$$\Rightarrow\lfloor\frac{{N}}{\mathrm{3}}\rfloor+\lfloor\frac{{N}}{\mathrm{35}}\rfloor=\lfloor\frac{{N}}{\mathrm{5}}\rfloor+\lfloor\frac{{N}}{\mathrm{7}}\rfloor \\ $$$${N}_{{min}} =\mathrm{5} \\ $$$${N}_{{max}} =\mathrm{65} \\ $$