anonymous
  • anonymous
Let a secret three digit number be cba. If the sum of cab + bac + bca + abc + abc = 2536, what is cba ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
abc+abc??
anonymous
  • anonymous
it was in the question..
anonymous
  • anonymous
We should let the number be secret.

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More answers

experimentX
  • experimentX
nice idea
anonymous
  • anonymous
no, open the secret :D
vishal_kothari
  • 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. ..
experimentX
  • experimentX
fun will be gone
anonymous
  • anonymous
hm.. i guess.. this is easier method: 2536 +/-multiple of 99=multiple of 111
perl
  • perl
vishal is a fraud , he finds answers online
perl
  • perl
here is the banana problem, compare http://www.gottfriedville.net/mathprob/misc-bananas.html
anonymous
  • anonymous
HAHA, lol
anonymous
  • anonymous
copy, paste?
anonymous
  • anonymous
or typed entirely?
perl
  • perl
he copied and pasted the whole thing. he is a fraud
anonymous
  • anonymous
ok, lets banish him xD
perl
  • perl
whats the challenge in googling?
anonymous
  • anonymous
no challenge, probably the question-poster is making a fun of us :/
perl
  • perl
well the question is fine, i dont like people who just post the whole answer. i want to work on it
perl
  • perl
well, maybe its not a big deal. ok , forget about it
perl
  • perl
better have the answer than not, right?
experimentX
  • experimentX
i agree ...
perl
  • perl
ok im angry that he gets a medal for it
perl
  • perl
a medal for using google?
experimentX
  • experimentX
it's all right ... medal's just nothing
anonymous
  • anonymous
but i gave him a medal, can i take it back? xD
anonymous
  • anonymous
i became his fan..
experimentX
  • experimentX
still ... i got to see a nice question and answer.
anonymous
  • anonymous
un-fan :D
perl
  • perl
hehe
experimentX
  • 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
anonymous
  • anonymous
but he can do it himself too..
perl
  • perl
this one is do-able
perl
  • perl
he says, i have all the answers on a file. yeah its called GOOGLE!!!
anonymous
  • anonymous
:D
anonymous
  • anonymous
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?
perl
  • perl
i still dont understand the gorilla problem
anonymous
  • anonymous
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
anonymous
  • anonymous
the gorilla cant understand how he can solve that too xD
anonymous
  • anonymous
213c +(a+b)221 = 2536 but this gives more then one solution if 221 is more than once
anonymous
  • anonymous
and it does
anonymous
  • anonymous
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...
anonymous
  • anonymous
\[cab+bac+bca+2abc=2536\]\[100∗(2a+2b+c)+10∗(2a+2b+c)+(a+b+3c)=2536\]\[a+b+3c=10x+3\tag1\]\[x+2a+2b+c=10y+6\tag2\]\[y+2a+2b+c=25\tag3\] 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
anonymous
  • anonymous
No :/ this isn't right

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