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Question Number 216076 by efronzo1 last updated on 27/Jan/25

((⌊(x/3) ⌋)/(⌊ (x/4) ⌋)) = ((21)/(16)) ; x=?

$$\:\:\:\frac{\lfloor\frac{\mathrm{x}}{\mathrm{3}}\:\rfloor}{\lfloor\:\frac{\mathrm{x}}{\mathrm{4}}\:\rfloor}\:=\:\frac{\mathrm{21}}{\mathrm{16}}\:;\:\mathrm{x}=? \\ $$

Answered by mr W last updated on 28/Jan/25

⌊(x/3)⌋=21k ⌊(x/4)⌋=16k ≠0 ⇒k≠0 (x/3)=21k+a with 0≤a<1 ⇒x=63k+3a (x/4)=16k+b with 0≤b<1 ⇒x=64k+4b 63k+3a=64k+4b −4<k=3a−4b<3 ⇒−3≤k≤2 ∧ k≠0 21k≤(x/3)<21k+1 3×21k≤x<3×21k+3 16k≤(x/4)<16k+1 4×16k≤x<4×16k+4 k=−3: x∈[−189, −186) x∈[−192, −188) ⇒−189≤x<−188 ✓ k=−2: x∈[−126, −123) x∈[−128, −124) ⇒−126≤x<−124 ✓ k=−1: x∈[−63, −60) x∈[−64, −60) ⇒−63≤x<−60 ✓ k=1: x∈[63, 66) x∈[64, 68) ⇒64≤x<66 ✓ k=2: x∈[126, 129) x∈[128, 132) ⇒128≤x<129 ✓

$$\lfloor\frac{{x}}{\mathrm{3}}\rfloor=\mathrm{21}{k} \\ $$$$\lfloor\frac{{x}}{\mathrm{4}}\rfloor=\mathrm{16}{k}\:\neq\mathrm{0}\:\Rightarrow{k}\neq\mathrm{0} \\ $$$$\frac{{x}}{\mathrm{3}}=\mathrm{21}{k}+{a}\:{with}\:\mathrm{0}\leqslant{a}<\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{63}{k}+\mathrm{3}{a} \\ $$$$\frac{{x}}{\mathrm{4}}=\mathrm{16}{k}+{b}\:{with}\:\mathrm{0}\leqslant{b}<\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{64}{k}+\mathrm{4}{b} \\ $$$$\mathrm{63}{k}+\mathrm{3}{a}=\mathrm{64}{k}+\mathrm{4}{b} \\ $$$$−\mathrm{4}<{k}=\mathrm{3}{a}−\mathrm{4}{b}<\mathrm{3} \\ $$$$\Rightarrow−\mathrm{3}\leqslant{k}\leqslant\mathrm{2}\:\wedge\:{k}\neq\mathrm{0} \\ $$$$\mathrm{21}{k}\leqslant\frac{{x}}{\mathrm{3}}<\mathrm{21}{k}+\mathrm{1} \\ $$$$\mathrm{3}×\mathrm{21}{k}\leqslant{x}<\mathrm{3}×\mathrm{21}{k}+\mathrm{3} \\ $$$$\mathrm{16}{k}\leqslant\frac{{x}}{\mathrm{4}}<\mathrm{16}{k}+\mathrm{1} \\ $$$$\mathrm{4}×\mathrm{16}{k}\leqslant{x}<\mathrm{4}×\mathrm{16}{k}+\mathrm{4} \\ $$$${k}=−\mathrm{3}: \\ $$$${x}\in\left[−\mathrm{189},\:−\mathrm{186}\right) \\ $$$${x}\in\left[−\mathrm{192},\:−\mathrm{188}\right) \\ $$$$\Rightarrow−\mathrm{189}\leqslant{x}<−\mathrm{188}\:\checkmark \\ $$$${k}=−\mathrm{2}: \\ $$$${x}\in\left[−\mathrm{126},\:−\mathrm{123}\right) \\ $$$${x}\in\left[−\mathrm{128},\:−\mathrm{124}\right) \\ $$$$\Rightarrow−\mathrm{126}\leqslant{x}<−\mathrm{124}\:\checkmark \\ $$$${k}=−\mathrm{1}: \\ $$$${x}\in\left[−\mathrm{63},\:−\mathrm{60}\right) \\ $$$${x}\in\left[−\mathrm{64},\:−\mathrm{60}\right) \\ $$$$\Rightarrow−\mathrm{63}\leqslant{x}<−\mathrm{60}\:\checkmark \\ $$$${k}=\mathrm{1}: \\ $$$${x}\in\left[\mathrm{63},\:\mathrm{66}\right) \\ $$$${x}\in\left[\mathrm{64},\:\mathrm{68}\right) \\ $$$$\Rightarrow\mathrm{64}\leqslant{x}<\mathrm{66}\:\checkmark \\ $$$${k}=\mathrm{2}: \\ $$$${x}\in\left[\mathrm{126},\:\mathrm{129}\right) \\ $$$${x}\in\left[\mathrm{128},\:\mathrm{132}\right) \\ $$$$\Rightarrow\mathrm{128}\leqslant{x}<\mathrm{129}\:\checkmark \\ $$

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