## sritama 3 years ago Let a secret three digit number be cba. If the sum of cab + bac + bca + abc + abc = 2536, what is cba ?

1. Arnab09

abc+abc??

2. sritama

it was in the question..

3. Ishaan94

We should let the number be secret.

4. experimentX

nice idea

5. Arnab09

no, open the secret :D

6. vishal_kothari

The sum of all six numbers is 222c + 222b + 222a If u omit cba = (100c + 10b + a), what is left is N = 122c + 212b + 221a which is what we are given. Observing 221 = 13*17, we have N = 5c + 4b mod 13 and N = 3c + 8b mod 17 (u could have used factors of 122 or 212, but 2*61 and 2*2*53 are not nearly so convenient as 13 and 17.) For N = 2536, we have N mod 13 = 1 and N mod 17 = 3 So u have 1 = 5c + 4b mod 13 and 3 = 3c + 8b mod 17 Since b and c are digits in range 0-9, u can try each value of b for 0 through 9 . Compute 4 x b and 4 x b mod 13. Subtract that from 1 to (5c mod 13). Then find a suitable value for c which satisfies that. For example 5c = 1 mod 13 -> 5c = 40, c = 8. compute 3c+8b (mod 17) and find the one which equals 3.. we see it is the row where b = 7 and c = 5. So we know that the original number was 57a. To determine a, we take 2536 - 122*5 - 212*7 = 2536 - 610 - 1484 = 442 Dividing by 221, we get a = 2, so the original number was 572. We can verify that 527+725+752+257+275 = 2536. ..

7. experimentX

fun will be gone

8. Arnab09

hm.. i guess.. this is easier method: 2536 +/-multiple of 99=multiple of 111

9. perl

vishal is a fraud , he finds answers online

10. perl

here is the banana problem, compare http://www.gottfriedville.net/mathprob/misc-bananas.html

11. Arnab09

HAHA, lol

12. Arnab09

copy, paste?

13. Arnab09

or typed entirely?

14. perl

he copied and pasted the whole thing. he is a fraud

15. Arnab09

ok, lets banish him xD

16. perl

whats the challenge in googling?

17. Arnab09

no challenge, probably the question-poster is making a fun of us :/

18. perl

well the question is fine, i dont like people who just post the whole answer. i want to work on it

19. perl

well, maybe its not a big deal. ok , forget about it

20. perl

better have the answer than not, right?

21. experimentX

i agree ...

22. perl

ok im angry that he gets a medal for it

23. perl

a medal for using google?

24. experimentX

it's all right ... medal's just nothing

25. Arnab09

but i gave him a medal, can i take it back? xD

26. sritama

i became his fan..

27. experimentX

still ... i got to see a nice question and answer.

28. Arnab09

un-fan :D

29. perl

hehe

30. experimentX

lol ... seriously i don't like the word fan ... i believe we all have equal abilities .. some use and some don't use ... but i think it's fun

31. Arnab09

but he can do it himself too..

32. perl

this one is do-able

33. perl

he says, i have all the answers on a file. yeah its called GOOGLE!!!

34. sritama

:D

35. myko

cab + bac + bca + abc + abc = 2536 100c+10a+b+100c+10a+c+100b+10c+a+2(100a+10b+c) = 2536 213c +221b + 221a =2536 maybe this is good start?

36. perl

i still dont understand the gorilla problem

37. Arnab09

ok, if u would replace last abc by cba, the result would be different, the sum would b divisible by 222 and the difference between them is divisible by 99 then, i guess hit and trial method

38. Arnab09

the gorilla cant understand how he can solve that too xD

39. myko

213c +(a+b)221 = 2536 but this gives more then one solution if 221 is more than once

40. myko

and it does

41. myko

cab + bac + bca + abc + abc = 2536 100c+10a+b+100c+10a+c+100b+10c+a+2(100a+10b+c) = 2536 213c +221b + 221a =2536 maybe i made a mistake here somehere...

42. Ishaan94

\[cab+bac+bca+2abc=2536\]\[100∗(2a+2b+c)+10∗(2a+2b+c)+(a+b+3c)=2536\]\[a+b+3c=10x+3\tag1\]\[x+2a+2b+c=10y+6\tag2\]\[y+2a+2b+c=25\tag3\] From (1), (2) and (3), \[x−11y=−22\] I think it would be safe to assume x=0 and y=2.\[2a+2b+c=23,a+b+3c=6\]We all know 0<a,b,c<9. So, it wouldn't be that hard to guess all the numbers.

43. Ishaan94

No :/ this isn't right