Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

abc+abc??
it was in the question..
We should let the number be secret.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

nice idea
no, open the secret :D
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. ..
fun will be gone
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
HAHA, lol
copy, paste?
or typed entirely?
he copied and pasted the whole thing. he is a fraud
ok, lets banish him xD
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?
i agree ...
ok im angry that he gets a medal for it
a medal for using google?
it's all right ... medal's just nothing
but i gave him a medal, can i take it back? xD
i became his fan..
still ... i got to see a nice question and answer.
un-fan :D
hehe
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..
this one is do-able
he says, i have all the answers on a file. yeah its called GOOGLE!!!
:D
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
and it does
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
No :/ this isn't right

Not the answer you are looking for?

Search for more explanations.

Ask your own question