Question-112573

Question Number 112573 by Aina Samuel Temidayo last updated on 08/Sep/20

Answered by Rasheed.Sindhi last updated on 08/Sep/20

Let two numbers are 10a+b & 10c+d with 10a+b>10c+d After written beside: 100(10a+b)+(10c+d) Absolute differnce (10a+b)−(10c+d) The equation will be 100(10a+b)+(10c+d) −{(10a+b)−(10c+d)}=5481 To determine: (10a+b)+(10c+d) 1000a+100b+10c+d −10a−b+10c+d=5481 ⇒990a+99b+20c+2d=5481 ⇒99(10a+b)+2(10c+d)=5481 10a+b=((5481−2(10c+d))/(99)) (10a+b)+(10c+d) =((5481−2(10c+d))/(99))+10c+d =((5481−2(10c+d)+990c+99d)/(99)) =((5481+970c+97d)/(99)) [We have to determine such values of c & d that ((5481+970c+97d)/(99)) ∈N]] Now, 1≤c≤9 ∧ 0≤d≤9: So, For c=1 & d=8: =((5481+970(1)+97(8))/(99))=73 (The numbers are 55 & 18)

$${Let}\:{two}\:{numbers}\:{are}\:\mathrm{10}{a}+{b}\:\& \\ $$$$\mathrm{10}{c}+{d}\:{with}\:\mathrm{10}{a}+{b}>\mathrm{10}{c}+{d} \\ $$$${After}\:{written}\:{beside}: \\ $$$$\mathrm{100}\left(\mathrm{10}{a}+{b}\right)+\left(\mathrm{10}{c}+{d}\right) \\ $$$${Absolute}\:{differnce} \\ $$$$\:\:\:\left(\mathrm{10}{a}+{b}\right)−\left(\mathrm{10}{c}+{d}\right) \\ $$$${The}\:{equation}\:{will}\:{be} \\ $$$$\:\mathrm{100}\left(\mathrm{10}{a}+{b}\right)+\left(\mathrm{10}{c}+{d}\right) \\ $$$$\:\:\:\:\:\:−\left\{\left(\mathrm{10}{a}+{b}\right)−\left(\mathrm{10}{c}+{d}\right)\right\}=\mathrm{5481} \\ $$$${To}\:{determine}: \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{10}{a}+{b}\right)+\left(\mathrm{10}{c}+{d}\right) \\ $$$$ \\ $$$$\mathrm{1000}{a}+\mathrm{100}{b}+\mathrm{10}{c}+{d} \\ $$$$\:\:\:\:\:−\mathrm{10}{a}−{b}+\mathrm{10}{c}+{d}=\mathrm{5481} \\ $$$$\Rightarrow\mathrm{990}{a}+\mathrm{99}{b}+\mathrm{20}{c}+\mathrm{2}{d}=\mathrm{5481} \\ $$$$\Rightarrow\mathrm{99}\left(\mathrm{10}{a}+{b}\right)+\mathrm{2}\left(\mathrm{10}{c}+{d}\right)=\mathrm{5481} \\ $$$$\:\:\:\:\:\mathrm{10}{a}+{b}=\frac{\mathrm{5481}−\mathrm{2}\left(\mathrm{10}{c}+{d}\right)}{\mathrm{99}} \\ $$$$\left(\mathrm{10}{a}+{b}\right)+\left(\mathrm{10}{c}+{d}\right) \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{5481}−\mathrm{2}\left(\mathrm{10}{c}+{d}\right)}{\mathrm{99}}+\mathrm{10}{c}+{d}\:\: \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{5481}−\mathrm{2}\left(\mathrm{10}{c}+{d}\right)+\mathrm{990}{c}+\mathrm{99}{d}}{\mathrm{99}}\:\: \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{5481}+\mathrm{970}{c}+\mathrm{97}{d}}{\mathrm{99}}\:\: \\ $$$$\:\left[{We}\:{have}\:{to}\:{determine}\:{such}\:{values}\right. \\ $$$$\left.{o}\left.{f}\:{c}\:\&\:{d}\:{that}\:\frac{\mathrm{5481}+\mathrm{970}{c}+\mathrm{97}{d}}{\mathrm{99}}\:\in\mathbb{N}\right]\right] \\ $$$${Now}, \\ $$$$\:\:\mathrm{1}\leqslant{c}\leqslant\mathrm{9}\:\wedge\:\mathrm{0}\leqslant{d}\leqslant\mathrm{9}: \\ $$$${So}, \\ $$$${For}\:{c}=\mathrm{1}\:\&\:{d}=\mathrm{8}: \\ $$$$\:\:\:\:\:=\frac{\mathrm{5481}+\mathrm{970}\left(\mathrm{1}\right)+\mathrm{97}\left(\mathrm{8}\right)}{\mathrm{99}}=\mathrm{73} \\ $$$$\left({The}\:{numbers}\:{are}\:\mathrm{55}\:\&\:\mathrm{18}\right) \\ $$

Commented by Rasheed.Sindhi last updated on 08/Sep/20