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sritama
Let a secret three digit number be cba. If the sum of cab + bac + bca + abc + abc = 2536, what is cba ?
it was in the question..
We should let the number be secret.
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. ..
hm.. i guess.. this is easier method: 2536 +/-multiple of 99=multiple of 111
vishal is a fraud , he finds answers online
here is the banana problem, compare http://www.gottfriedville.net/mathprob/misc-bananas.html
he copied and pasted the whole thing. he is a fraud
whats the challenge in googling?
no challenge, probably the question-poster is making a fun of us :/
well the question is fine, i dont like people who just post the whole answer. i want to work on it
well, maybe its not a big deal. ok , forget about it
better have the answer than not, right?
ok im angry that he gets a medal for it
it's all right ... medal's just nothing
but i gave him a medal, can i take it back? xD
still ... i got to see a nice question and answer.
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
but he can do it himself too..
he says, i have all the answers on a file. yeah its called GOOGLE!!!
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?
i still dont understand the gorilla problem
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
the gorilla cant understand how he can solve that too xD
213c +(a+b)221 = 2536 but this gives more then one solution if 221 is more than once
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...
\[cab+bac+bca+2abc=2536\]\[100∗(2a+2b+c)+10∗(2a+2b+c)+(a+b+3c)=2536\]\[a+b+3c=10x+3\tag1\]\[x+2a+2b+c=10y+6\tag2\]\[y+2a+2b+c=25\tag3\] 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.
No :/ this isn't right