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Question Number 90698 by jagoll last updated on 25/Apr/20

how many solution the equation ⌊ x ⌋ +2016. {x} = 38?

$${how}\:{many}\:{solution}\:{the}\:{equation} \\ $$$$\lfloor\:{x}\:\rfloor\:+\mathrm{2016}.\:\left\{{x}\right\}\:=\:\mathrm{38}? \\ $$

Commented by mr W last updated on 25/Apr/20

in general ⌊ x ⌋ +n{x} = m has n solutions.

$${in}\:{general} \\ $$$$\lfloor\:{x}\:\rfloor\:+{n}\left\{{x}\right\}\:=\:{m} \\ $$$${has}\:{n}\:{solutions}. \\ $$

Commented by jagoll last updated on 25/Apr/20

if change n ⌊x ⌋ + {x} = m how sir?

$${if}\:{change}\:{n}\:\lfloor{x}\:\rfloor\:+\:\left\{{x}\right\}\:=\:{m} \\ $$$${how}\:{sir}? \\ $$

Commented by mr W last updated on 25/Apr/20

it has not much sense. {x} must be zero. i.e. x must be integer. ⌊x⌋=x nx=m if m≡0 mod (n) then one solution: x=(m/n) otherwise no solution.

$${it}\:{has}\:{not}\:{much}\:{sense}. \\ $$$$\left\{{x}\right\}\:{must}\:{be}\:{zero}.\:{i}.{e}.\:{x}\:{must}\:{be}\:{integer}. \\ $$$$\lfloor{x}\rfloor={x} \\ $$$${nx}={m} \\ $$$${if}\:\:{m}\equiv\mathrm{0}\:{mod}\:\left({n}\right)\:{then}\:{one}\:{solution}: \\ $$$${x}=\frac{{m}}{{n}} \\ $$$${otherwise}\:{no}\:{solution}. \\ $$

Answered by mr W last updated on 25/Apr/20

x=⌊x⌋+{x}=n+f n+2016f=38 0≤f=((38−n)/(2016))<1 ⇒0≤38−n<2016 ⇒−1978<n≤38 ⇒−1977≤n≤38 i.e. there are 2016 solutions.

$${x}=\lfloor{x}\rfloor+\left\{{x}\right\}={n}+{f} \\ $$$${n}+\mathrm{2016}{f}=\mathrm{38} \\ $$$$\mathrm{0}\leqslant{f}=\frac{\mathrm{38}−{n}}{\mathrm{2016}}<\mathrm{1} \\ $$$$\Rightarrow\mathrm{0}\leqslant\mathrm{38}−{n}<\mathrm{2016} \\ $$$$\Rightarrow−\mathrm{1978}<{n}\leqslant\mathrm{38} \\ $$$$\Rightarrow−\mathrm{1977}\leqslant{n}\leqslant\mathrm{38} \\ $$$${i}.{e}.\:{there}\:{are}\:\mathrm{2016}\:{solutions}. \\ $$

Commented by jagoll last updated on 25/Apr/20

why sir has 2016 solution?

$${why}\:{sir}\:{has}\:\mathrm{2016}\:{solution}? \\ $$

Commented by mr W last updated on 25/Apr/20

for each n ∈[−1977, 38] there is a solution x=n+((38−n)/(2016)). we have totally 2016 values of n: −1977, −1976, ...,−1, 0, 1, ..., 38 therefore there are 2016 solutions for ⌊x⌋+2016{x}=38.

$${for}\:{each}\:{n}\:\in\left[−\mathrm{1977},\:\mathrm{38}\right]\:{there}\:{is}\:{a} \\ $$$${solution}\:{x}={n}+\frac{\mathrm{38}−{n}}{\mathrm{2016}}. \\ $$$${we}\:{have}\:{totally}\:\mathrm{2016}\:{values}\:{of}\:{n}: \\ $$$$−\mathrm{1977},\:−\mathrm{1976},\:...,−\mathrm{1},\:\mathrm{0},\:\mathrm{1},\:...,\:\mathrm{38} \\ $$$${therefore}\:{there}\:{are}\:\mathrm{2016}\:{solutions} \\ $$$${for}\:\lfloor{x}\rfloor+\mathrm{2016}\left\{{x}\right\}=\mathrm{38}. \\ $$

Commented by jagoll last updated on 25/Apr/20

oo it does mean −1977,−1976,..., 38 equal to AP with d = 1 and T_1 = −1977 so 38 = −1977 +(n−1)×1 38 = n −1978 n = 38 + 1978 = 2016 sir?

$${oo}\:{it}\:{does}\:{mean}\: \\ $$$$−\mathrm{1977},−\mathrm{1976},...,\:\mathrm{38}\:{equal}\:{to} \\ $$$${AP}\:{with}\:{d}\:=\:\mathrm{1}\:{and}\:{T}_{\mathrm{1}} \:=\:−\mathrm{1977} \\ $$$${so}\:\mathrm{38}\:=\:−\mathrm{1977}\:+\left({n}−\mathrm{1}\right)×\mathrm{1}\: \\ $$$$\mathrm{38}\:=\:{n}\:−\mathrm{1978}\: \\ $$$${n}\:=\:\mathrm{38}\:+\:\mathrm{1978}\:=\:\mathrm{2016}\:{sir}? \\ $$

Commented by mr W last updated on 25/Apr/20

yes. from −1977 to 38 there are 2016 integers.

$${yes}. \\ $$$${from}\:−\mathrm{1977}\:{to}\:\mathrm{38}\:{there}\:{are}\:\mathrm{2016} \\ $$$${integers}. \\ $$

Commented by jagoll last updated on 25/Apr/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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