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 substract 1089 from the desired number we obtain a number containing the same digits as the desired number but in reverse order. b)Prove that if the sum k+m+n of three natural numbers is divisible by 6 then k^3 +m^3 +n^3 is also divisible by6


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

$a) Find a four - digit number that$ $satisfies the following condition :$ $the sum of the squares of the extreme digits$ $is equal to 13; the sum of the squares of$ $the middle digits is equal to 85; if we$ $substract 1089 from the desired$ $number we obtain a number containing$ $the same digits as the desired number$ $but in reverse order .$ $b) Prove that if the sum k + m + n of$ $three natural numbers is divisible by$ $6 then k^{3} + m^{3} + n^{3} is also divisible by 6$
Commented by Rasheed.Sindhi last updated on 18/Sep/20

$C o u n t e r e x a m p l e$ $6 ∣ (1 + 2 + 3) b u t 6 ∤ (1^{2} + 2^{2} + 3^{2})$
Commented by 1549442205PVT last updated on 18/Sep/20

$Thank Sir . Exuse me, typo . I corrected$
	Answered by mr W last updated on 18/Sep/20

	$[a b c d]$ $a^{2} + d^{2} = 13 \Rightarrow (a, d) = (2, 3) o r (3, 2)$ $b^{2} + c^{2} = 85 \Rightarrow (b, c) = (9, 2) o r (2, 9) o r (6, 7) o r (7, 6)$ $[a b c d] - 1089 = [d c b a]$ $\Rightarrow a > d \Rightarrow a = 3, d = 2$ $[3 b c 2] - 1089 = [2 c b 3]$ $3000 + 100 b + 10 c + 2 - 1089 = 2000 + 100 c + 10 b + 3$ $\Rightarrow b = c + 1 \Rightarrow b = 7, c = 6$ $\Rightarrow t h e n u m b e r i s 3762 .$
	Commented by Rasheed.Sindhi last updated on 18/Sep/20

	${Y o u}^{'} r e f a s t e r s i r!$
	Commented by mr W last updated on 18/Sep/20

	$s o r r y s i r! i g i v e t o o l e s s e x p l a n a t i o n .$
	Commented by mr W last updated on 18/Sep/20

	$c a n y o u c o m f i r m Q 113710 s i r ?$
	Commented by Rasheed.Sindhi last updated on 18/Sep/20

	$S o r r y s i r I {c a n}^{'} t .$
	Commented by Rasheed.Sindhi last updated on 18/Sep/20

	$I t h i n k y o u r a n s w e r i s m o r e$ $a l g e b r a i c - b a s e d . W h e r e a s m y$ $a n s w e r s o m e w h a t m o r e d e p e n d s o n$ $t r i a l . S o I l i k e y o u r s!$
	Commented by mr W last updated on 18/Sep/20

	$t h a n k s s i r!$
	Answered by Rasheed.Sindhi last updated on 18/Sep/20
	$Reauired number: abcd a^2 +d^2 =13.......(i) b^2 +c^2 =85.........(ii) abcd−1089=dcba.....(iii) dcba<abcd⇒d<a a^2 +d^2 =13 ∧ d<a⇒a^2 =9∧d^2 =4 ⇒a=3 ∧ d=2 The number is now 3bc2 b^2 +c^2 =85 85 is the sum of 81 and 4 (square or 36 and 49 ((square numbers) So {b,c}={2,9} or {6,7} The number may be 3922 or 3292 or 3672 or 3762 Applying condition(iii) The number is 3762$
	$R e a u i r e d n u m b e r : a b c d$ $a^{2} + d^{2} = 13 . . . . . . . (i)$ $b^{2} + c^{2} = 85 . . . . . . . . . (i i)$ $a b c d - 1089 = d c b a . . . . . (i i i)$ $d c b a < a b c d \Rightarrow d < a$ $a^{2} + d^{2} = 13 \land d < a \Rightarrow a^{2} = 9 \land d^{2} = 4$ $\Rightarrow a = 3 \land d = 2$ $T h e n u m b e r i s n o w 3 b c 2$ $b^{2} + c^{2} = 85$ $85 i s t h e s u m o f 81 a n d 4 (s q u a r e$ $o r 36 a n d 49 ((s q u a r e n u m b e r s)$ $S o {b, c} = {2, 9} o r {6, 7}$ $T h e n u m b e r m a y b e$ $3922 o r 3292 o r 3672 o r 3762$ $A p p l y i n g c o n d i t i o n (i i i)$ $T h e n u m b e r i s 3762$
	Answered by MJS_new last updated on 18/Sep/20

	$6 ∣ (k + m + n)$ $k + m + n = 6 x$ $k = 6 x - m - n$ $k^{3} + m^{3} + n^{3} =$ $= 18 x (12 x^{2} - 6 (m + n) x + {(m + n)}^{2}) - 3 m n (m + n)$ $6 ∣ (18 x (12 x^{2} - 6 (m + n) x + {(m + n)}^{2}))$ $we need to show 6 ∣ (3 m n (m + n))$ $\Leftrightarrow 2 ∣ m n (m + n)$ $easy to see this is true if 2 ∣ m \lor 2 ∣ n$ $and if both m, n are uneven 2 ∣ (m + n)$ $\Rightarrow proved$
	Answered by 1549442205PVT last updated on 20/Sep/20
	$Thank all Sirs a)Denote the number we need find by mnpq^(−) .From the hypothesis we have m^2 +q^2 =13=3^2 +2^2 ⇒mq^(−) ∈{32,23} n^2 +q^2 =85=9^2 +2^2 =7^2 +6^2 ⇒np^(−) ∈{92,29,76,67} From the condition mnpq^(−) −1089=qpnm^(−) ⇔mnpq^(−) −qpnm^(−) =1089⇒m=3,q=2 we hav mnpq^(−) ∈{3922,3292,3762,3672} Hence,3np2^(−) −2pn3^(−) =1089.This gives us np^(−) =76⇒mnpq^(−) =3762 b)Applying the identity a^3 +b^3 +c^3 −3abc=(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca) we have: k^3 +m^3 +n^3 =3mnk+(k+m+n)(k^2 +m^2 +n^2 −km−kn−mn) From the hypothesis m+n+k⋮6 ⇒m+n+k=6p(p∈Z)⇒among three numbers m,n,k at least one number is even (since if all three numbers are odd then m+n+k is odd then m+n+k isn′t divisible by 2,so isn′t divisible by 6) ⇒3mnk⋮6,but m+n+k⋮6 ⇒m^3 +n^3 +k^3 ⋮6(Q.E.D)$
	$Thank all Sirs$ $a) Denote the number we need find by$ $\overset{―}{mnpq} . From the hypothesis we have$ $m^{2} + q^{2} = 13 = 3^{2} + 2^{2} \Rightarrow \overset{―}{mq} \in {32, 23}$ $n^{2} + q^{2} = 85 = 9^{2} + 2^{2} = 7^{2} + 6^{2} \Rightarrow \overset{―}{np} \in {92, 29, 76, 67}$ $From the condition \overset{―}{mnpq} - 1089 = \overset{―}{qpnm}$ $\Leftrightarrow \overset{―}{mnpq} - \overset{―}{qpnm} = 1089 \Rightarrow m = 3, q = 2$ $we hav \overset{―}{mnpq} \in {3922, 3292, 3762, 3672}$ $Hence, \overset{―}{3 np 2} - \overset{―}{2 pn 3} = 1089 . This gives$ $us \overset{―}{np} = 76 \Rightarrow \overset{―}{mnpq} = 3762$ $b) Applying the identity$ $a^{3} + b^{3} + c^{3} - 3 abc = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ca)$ $we have :$ $k^{3} + m^{3} + n^{3} = 3 mnk + (k + m + n) (k^{2} + m^{2}$ $+ n^{2} - km - kn - mn)$ $From the hypothesis m + n + k ⋮ 6$ $\Rightarrow m + n + k = 6 p (p \in Z) \Rightarrow among three$ $numbers m, n, k at least one number$ $is even (since if all three numbers$ $are odd then m + n + k is odd then$ $m + n + k {isn}^{'} t divisible by 2, so {isn}^{'} t$ $divisible by 6)$ $\Rightarrow 3 mnk ⋮ 6, but m + n + k ⋮ 6$ $\Rightarrow m^{3} + n^{3} + k^{3} ⋮ 6 (Q . E . D)$
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