a-Find-a-four-digit-number-that-satisfies-the-following-condition-the-sum-of-the-squares-of-the-extreme-digits-is-equal-to-13-the-sum-of-the-squares-of-the-middle-digits-is-equal-to-85-if-we-substrac

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Question Number 114326 by 1549442205PVT last updated on 18/Sep/20

$$\mathrm{a})\mathrm{Find}\mathrm{a}\mathrm{four}-\mathrm{digit}\mathrm{number}\mathrm{that}$$$$\mathrm{satisfies}\mathrm{the}\mathrm{following}\mathrm{condition}:$$$$\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{squares}\mathrm{of}\mathrm{the}\mathrm{extreme}\mathrm{digits}$$$$\mathrm{is}\mathrm{equal}\mathrm{to}13;\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{squares}\mathrm{of}$$$$\mathrm{the}\mathrm{middle}\mathrm{digits}\mathrm{is}\mathrm{equal}\mathrm{to}85;\mathrm{if}\mathrm{we}$$$$\mathrm{substract}1089\mathrm{from}\mathrm{the}\mathrm{desired}$$$$\mathrm{number}\mathrm{we}\mathrm{obtain}\mathrm{a}\mathrm{number}\mathrm{containing}$$$$\mathrm{the}\mathrm{same}\mathrm{digits}\mathrm{as}\mathrm{the}\mathrm{desired}\mathrm{number}$$$$\mathrm{but}\mathrm{in}\mathrm{reverse}\mathrm{order}.$$$$\mathrm{b})\mathrm{Prove}\mathrm{that}\mathrm{if}\mathrm{the}\mathrm{sum}\mathrm{k}+\mathrm{m}+\mathrm{n}\mathrm{of}$$$$\mathrm{three}\mathrm{natural}\mathrm{numbers}\mathrm{is}\mathrm{divisible}\mathrm{by}$$$$6\mathrm{then}{\mathrm{k}}^3+{\mathrm{m}}^3+{\mathrm{n}}^3\mathrm{is}\mathrm{also}\mathrm{divisible}\mathrm{by}6$$

Commented by Rasheed.Sindhi last updated on 18/Sep/20

$$Counterexample$$$$6\mid (1+2+3)but6\nmid (1^2+2^2+3^2)$$

Commented by 1549442205PVT last updated on 18/Sep/20

$$\mathrm{Thank}\mathrm{Sir}.\mathrm{Exuse}\mathrm{me},\mathrm{typo}.\mathrm{I}\mathrm{corrected}$$

Answered by mr W last updated on 18/Sep/20

$$\left[abcd\right]$$$$a^2+d^2=13\Rightarrow (a,d)=(2,3)or(3,2)$$$$b^2+c^2=85\Rightarrow (b,c)=(9,2)or(2,9)or(6,7)or(7,6)$$$$\left[abcd\right]-1089=\left[dcba\right]$$$$\Rightarrow a>d\Rightarrow a=3,d=2$$$$\left[3bc2\right]-1089=\left[2cb3\right]$$$$3000+100b+10c+2-1089=2000+100c+10b+3$$$$\Rightarrow b=c+1\Rightarrow b=7,c=6$$$$\Rightarrow thenumberis3762.$$

Commented by Rasheed.Sindhi last updated on 18/Sep/20

$${You}^{\prime }refastersir!$$

Commented by mr W last updated on 18/Sep/20

$$sorrysir!igivetoolessexplanation.$$

Commented by mr W last updated on 18/Sep/20

$$canyoucomfirmQ113710sir?$$

Commented by Rasheed.Sindhi last updated on 18/Sep/20

$$SorrysirI{can}^{\prime }t.$$

Commented by Rasheed.Sindhi last updated on 18/Sep/20

$$Ithinkyouranswerismore$$$$algebraic-based.Whereasmy$$$$answersomewhatmoredependson$$$$trial.SoIlikeyours!$$

Commented by mr W last updated on 18/Sep/20

$$thankssir!$$

Answered by Rasheed.Sindhi last updated on 18/Sep/20

$$Reauirednumber:abcd$$$$a^2+d^2=13\dots \dots .\left(i\right)$$$$b^2+c^2=85\dots \dots \dots \left(ii\right)$$$$abcd-1089=dcba\dots ..\left(iii\right)$$$$dcba<abcd\Rightarrow d<a$$$$a^2+d^2=13\wedge d<a\Rightarrow a^2=9\wedge d^2=4$$$$\Rightarrow a=3\wedge d=2$$$$Thenumberisnow3bc2$$$$b^2+c^2=85$$$$85isthesumof81and4(square$$$$or36and49(\left(squarenumbers\right)$$$$So\{b,c\}=\{2,9\}or\{6,7\}$$$$Thenumbermaybe$$$$3922or3292or3672or3762$$$$Applyingcondition\left(iii\right)$$$$Thenumberis3762$$$$$$$$$$

Answered by MJS_new last updated on 18/Sep/20

$$6\mid (k+m+n)$$$$k+m+n=6x$$$$k=6x-m-n$$$$k^3+m^3+n^3=$$$$=18x(12x^2-6(m+n)x+{(m+n)}^2)-3mn(m+n)$$$$6\mid \left(18x(12x^2-6(m+n)x+{(m+n)}^2)\right)$$$$\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{show}6\mid \left(3mn(m+n)\right)$$$$\iff 2\mid mn(m+n)$$$$\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{this}\mathrm{is}\mathrm{true}\mathrm{if}2\mid m\vee 2\mid n$$$$\mathrm{and}\mathrm{if}\mathrm{both}m,n\mathrm{are}\mathrm{uneven}2\mid (m+n)$$$$\Rightarrow \mathrm{proved}$$

Answered by 1549442205PVT last updated on 20/Sep/20

$$\mathrm{Thank}\mathrm{all}\mathrm{Sirs}$$$$\mathrm{a})\mathrm{Denote}\mathrm{the}\mathrm{number}\mathrm{we}\mathrm{need}\mathrm{find}\mathrm{by}$$$$\overline{\mathrm{mnpq}}.\mathrm{From}\mathrm{the}\mathrm{hypothesis}\mathrm{we}\mathrm{have}$$$${\mathrm{m}}^2+{\mathrm{q}}^2=13=3^2+2^2\Rightarrow \overline{\mathrm{mq}}\in \{32,23\}$$$${\mathrm{n}}^2+{\mathrm{q}}^2=85=9^2+2^2=7^2+6^2\Rightarrow \overline{\mathrm{np}}\in \{92,29,76,67\}$$$$\mathrm{From}\mathrm{the}\mathrm{condition}\overline{\mathrm{mnpq}}-1089=\overline{\mathrm{qpnm}}$$$$\iff \overline{\mathrm{mnpq}}-\overline{\mathrm{qpnm}}=1089\Rightarrow \mathrm{m}=3,\mathrm{q}=2$$$$\mathrm{we}\mathrm{hav}\overline{\mathrm{mnpq}}\in \{3922,3292,3762,3672\}$$$$\mathrm{Hence},\overline{3\mathrm{np}2}-\overline{2\mathrm{pn}3}=1089.\mathrm{This}\mathrm{gives}$$$$\mathrm{us}\overline{\mathrm{np}}=76\Rightarrow \overline{\mathrm{mnpq}}=3762$$$$\mathrm{b})\mathrm{Applying}\mathrm{the}\mathrm{identity}$$$${\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3-3\mathrm{abc}=(\mathrm{a}+\mathrm{b}+\mathrm{c})({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca})$$$$\mathrm{we}\mathrm{have}:$$$${\mathrm{k}}^3+{\mathrm{m}}^3+{\mathrm{n}}^3=3\mathrm{mnk}+(\mathrm{k}+\mathrm{m}+\mathrm{n})({\mathrm{k}}^2+{\mathrm{m}}^2$$$$+{\mathrm{n}}^2-\mathrm{km}-\mathrm{kn}-\mathrm{mn})$$$$\mathrm{From}\mathrm{the}\mathrm{hypothesis}\mathrm{m}+\mathrm{n}+\mathrm{k}⋮6$$$$\Rightarrow \mathrm{m}+\mathrm{n}+\mathrm{k}=6\mathrm{p}(\mathrm{p}\in \mathrm{Z})\Rightarrow \mathrm{among}\mathrm{three}$$$$\mathrm{numbers}\mathrm{m},\mathrm{n},\mathrm{k}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{number}$$$$\mathrm{is}\mathrm{even}(\mathrm{since}\mathrm{if}\mathrm{all}\mathrm{three}\mathrm{numbers}$$$$\mathrm{are}\mathrm{odd}\mathrm{then}\mathrm{m}+\mathrm{n}+\mathrm{k}\mathrm{is}\mathrm{odd}\mathrm{then}$$$$\mathrm{m}+\mathrm{n}+\mathrm{k}{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{divisible}\mathrm{by}2,\mathrm{so}{\mathrm{isn}}^{\prime }\mathrm{t}$$$$\mathrm{divisible}\mathrm{by}6)$$$$\Rightarrow 3\mathrm{mnk}⋮6,\mathrm{but}\mathrm{m}+\mathrm{n}+\mathrm{k}⋮6$$$$\Rightarrow {\mathrm{m}}^3+{\mathrm{n}}^3+{\mathrm{k}}^3⋮6(\mathbf{Q}.\mathbf{E}.\mathbf{D})$$

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