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N-1335-in-base-a-Write-N-in-base-a-1-




Question Number 122679 by mathocean1 last updated on 18/Nov/20
N=1335 in base a.   Write N in base a+1
$${N}=\mathrm{1335}\:{in}\:{base}\:{a}.\: \\ $$$${Write}\:{N}\:{in}\:{base}\:{a}+\mathrm{1} \\ $$
Answered by mr W last updated on 18/Nov/20
N=(1335)_a   =1×a^3 +3×a^2 +3×a+5  =1×a^3 +3×a^2 +3×a+1+4  =(a+1)^3 +4  =1×(a+1)^3 +0×(a+1)^2 +0×(a+1)+4  =(1004)_(a+1)
$${N}=\left(\mathrm{1335}\right)_{{a}} \\ $$$$=\mathrm{1}×{a}^{\mathrm{3}} +\mathrm{3}×{a}^{\mathrm{2}} +\mathrm{3}×{a}+\mathrm{5} \\ $$$$=\mathrm{1}×{a}^{\mathrm{3}} +\mathrm{3}×{a}^{\mathrm{2}} +\mathrm{3}×{a}+\mathrm{1}+\mathrm{4} \\ $$$$=\left({a}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{4} \\ $$$$=\mathrm{1}×\left({a}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{0}×\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{0}×\left({a}+\mathrm{1}\right)+\mathrm{4} \\ $$$$=\left(\mathrm{1004}\right)_{{a}+\mathrm{1}} \\ $$
Answered by MJS_new last updated on 18/Nov/20
1004  a^3 +3a^2 +3a+5=α(a+1)^3 +β(a+1)^2 +γ(a+1)+δ  (1−α)a^3 +(3−3α−β)a^2 +(3−3α+2β−γ)a+(5−α−β−γ−δ)=0  all brackets must be zero ⇒  α=1∧β=0∧γ=0∧δ=4
$$\mathrm{1004} \\ $$$${a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{5}=\alpha\left({a}+\mathrm{1}\right)^{\mathrm{3}} +\beta\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\gamma\left({a}+\mathrm{1}\right)+\delta \\ $$$$\left(\mathrm{1}−\alpha\right){a}^{\mathrm{3}} +\left(\mathrm{3}−\mathrm{3}\alpha−\beta\right){a}^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{3}\alpha+\mathrm{2}\beta−\gamma\right){a}+\left(\mathrm{5}−\alpha−\beta−\gamma−\delta\right)=\mathrm{0} \\ $$$$\mathrm{all}\:\mathrm{brackets}\:\mathrm{must}\:\mathrm{be}\:\mathrm{zero}\:\Rightarrow \\ $$$$\alpha=\mathrm{1}\wedge\beta=\mathrm{0}\wedge\gamma=\mathrm{0}\wedge\delta=\mathrm{4} \\ $$

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