Thursday, 12 May 2016

number theory - Solutions to 100a+10b+c=11a2+11b2+11c2 and 100a+10b+c=11a2+11b2+11c2+k



The problem states that




Find all three digit natural numbers such that when the number is divided by 11 gives the quotient equal to sum of squares of their digits





Since there is no information about whether remainder is 0 or not , I firstly assumed that the question is talking about the numbers perfectly divisible by 11



Now I have 80 numbers left , I can check them separately , but it will be lengthy



I made an equation 100a+10b+c=11a2+11b2+11c2



Rearranged and got a(11a100)+b(11b10)+c(11c1)=0



I have 10 values for a,b,c={1,2,3,4,5,6,7,8,9,0}




I made a table corresponding to a(11a100),b(11b10),c(11c1) and found that only for a=b=5,c=0 and a=8,b=0,c=3 are giving their sum 0 , hence 550 and 803 are the only numbers satisfying given property and divisible by 11.



Now I have two questions:




1.)Is my way and my answer correct? If no, then where have I misunderstood?



2.)What about the numbers which are not divisible by 11?



As mentioned by an answer er of this post , there are six such numbers which are not divisible by 11 , but still give the quotient the sum of square of their digits. But answer er found it using a computer program, which is not suitable for pen paper mathematics. So how can I find those six numbers ?




Answer



We want to find all sets of integers a,b,c,k such that 100a+10b+c=11a2+11b2+11c2+k
where 1a9,0b9,0c9 and 0k10.



We have
(1)k+b(11b10)+c(11c1)=a(10011a)



Since a(10011a)5(10011×5)=225, we have
k+b(11b10)+c(11c1)225




Since we have that b(11b10)6(11×610)=336>225 for b6 and that c(11c1)5(11×51)=270>225 for c5, from (2), we have to have b5 and c4.



Also, from (1) with 0k10, solving 0100a+10b+c11a211b211c210for a givesa[50m11,50m11011][50+m11011,50+m11]where m=250011(11b2+11c210bc).



Now, we see that 50m11150m110111521m1631fora=150m11250m11011784m894fora=250m11350m11011289m399fora=350m11450m11011110m146fora=450+m11011550+m11100m125fora=550+m11011650+m11256m356fora=650+m11011750+m11729m829fora=750+m11011850+m111444m1544fora=850+m11011950+m112401m2501fora=9



Case 1 : When b=5, since b(11b10)=225 with (2), we have to have k=c(11c1)=0,a=5 giving a=5,b=5,c=0,k=0.



Case 2 : When b=4, since b(11b10)=136,

(2)k+c(11c1)225136=89from which we have to have c2 since c(11c1)3×(11×31)=96>89 for c3.




  • Case 2-1 : For c=0, we have m=1004. So, there is no integer a satisfying (3).


  • Case 2-2 : For c=1, we have m=894. So, we have a=2, and k=10 from (1).


  • Case 2-3 : For c=2, we have m=542. So, there is no integer a satisfying (3).




Case 3 : When b=3, since b(11b10)=69,
(2)k+c(11c1)22569=156from which we have to have c3 since c(11c1)4×(11×41)=172>156 for c4.





  • Case 3-1 : For c=0, we have m=1741. So, there is no integer a satisfying (3).


  • Case 3-2 : For c=1, we have m=1631. So, we have a=1, and k=10 from (1).


  • Case 3-3 : For c=2, we have m=1279. So, there is no integer a satisfying (3).


  • Case 3-4 : For c=3, we have m=685. So, there is no integer a satisfying (3).




Case 4 : When b=2, we have c4.





  • Case 4-1 : For c=0, we have m=2236. So, there is no integer a satisfying (3).


  • Case 4-2 : For c=1, we have m=2126. So, there is no integer a satisfying (3).


  • Case 4-3 : For c=2, we have m=1774. So, there is no integer a satisfying (3).


  • Case 4-4 : For c=3, we have m=1180. So, there is no integer a satisfying (3).


  • Case 4-5 : For c=4, we have m=344. So, we have a=3,6, and (a,b,c,k)=(3,2,4,5),(6,2,4,8) from (1).




Case 5 : When b=1, we have c4.





  • Case 5-1 : For c=0, we have m=2489. So, we have a=9 and k=8 from (1).


  • Case 5-2 : For c=1, we have m=2379. So, there is no integer a satisfying (3).


  • Case 5-3 : For c=2, we have m=2027. So, there is no integer a satisfying (3).


  • Case 5-4 : For c=3, we have m=1433. So, there is no integer a satisfying (3).


  • Case 5-5 : For c=4, we have m=597. So, there is no integer a satisfying (3).




Case 6 : When b=0, we have c4.





  • Case 6-1 : For c=0, we have m=2500. So, we have a=9, and k=9 from (1).


  • Case 6-2 : For c=1, we have m=2390. So, there is no integer a satisfying (3).


  • Case 6-3 : For c=2, we have m=2038. So, there is no integer a satisfying (3).


  • Case 6-4 : For c=3, we have m=1444. So, we have a=8, and k=0 from (1).


  • Case 6-5 : For c=4, we have m=608. So, there is no integer a satisfying (3).




Therefore, the answer is
(a,b,c,k)= (5,5,0,0),(2,4,1,10),(1,3,1,10),(3,2,4,5)(6,2,4,8),(9,1,0,8),(9,0,0,9),(8,0,3,0)



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