anonymous
  • anonymous
Fooool's problems of the day,two *very very* easy problem: 1) How many four digit numbers of the form AABA are divisible by \(33\)? 2) How many three digit even numbers are there such that if they are subtracted from their reverse the result is a positive multiple of \(66\)? Lets see, who can come up with the solution first!
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
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

anonymous
  • anonymous
1. 0?
King
  • King
no i found one for the 1st one its not 0....wait...
anonymous
  • anonymous
I am sorry Ishaan but that's not the right answer.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[\frac{{AA*100} + BA}{3*11} = \frac{AA*100}{11*3} + \frac{BA}{11*3}\] I don't think there is a number in the form of BA divisible by 11. BTW What do you mean by AABA, numbers like 1121, 1131, 2242?
anonymous
  • anonymous
\( 3333 \) is a valid isn't? ;)
anonymous
  • anonymous
But you said AABA? Hmm if B=A is valid then, {3333, 6666, 9999} are clear solutions.
anonymous
  • anonymous
I never said \(B \neq A \) ;)
Mr.Math
  • Mr.Math
1) All numbers in the form \(AABA\) such that: i) \(3A+B=k\), where k is a number divisible by 3, and ii) \(B-A=c\), where c is divisible by 11. Now, for k, we have \(0\le k \le 36\). There are \(\frac{36}{3}+1=13\) such k in this interval. [That's {0,3,6,..36}] For c, we have \(0\le c \le 9\), there's only one c that's divisible by 11 which c=0. Hence \[B-A=0 \implies B=A\] and \[3A+B=0, 3, 6,.. 36\] From here we can conclude that the numbers are {0000, 3333, 6666, 9999}.
King
  • King
answer is 3?
Mr.Math
  • Mr.Math
4.
King
  • King
0000 is 0 not a 4 digit no. ...........
anonymous
  • anonymous
Hmm then Zero? //Now it's confusing me, I am moving on to Second part. Let the three Digit even number be ABC ABC - CBA Now for even ABC, C has to even A*100 + B*10 + C - C*100 -B*10 -A = 66k A(100 - 1) - C (100- 1) = 66k 3*33(A-C) = 2*33*k \[\frac{3}{2} * (A-C) = k\] We know C is even and to satisfy k as an Integer A has to be even too (two in the denominator) So, all the numbers in the form of ABC which have A and C as even numbers.
Mr.Math
  • Mr.Math
Hmm, I need to check the definition of a digit. But the idea is clear, I think.
King
  • King
@ishaan dont we have to find the amnt?
anonymous
  • anonymous
0000 is not a four digit number.
King
  • King
yes!!!
Mr.Math
  • Mr.Math
Then 3 such numbers exist.
anonymous
  • anonymous
Try the second one now :)
King
  • King
i fnd 1 number fr the 2nd one..... 400
Mr.Math
  • Mr.Math
I think Ishaan already did it.
King
  • King
but then we hav o find the amnt
King
  • King
is 0 div by 66?
Mr.Math
  • Mr.Math
I would rather let him find it, or I would be stealing his work :P
anonymous
  • anonymous
Ishaan's answer is probably 4*10*5 but that doesn't seem to be the correct answer.
King
  • King
is 0 div. by 66.....
anonymous
  • anonymous
Oh, I wasn't trying to find the number, Limit of laziness. Let me try now!
NotTim
  • NotTim
How do you learn how to solve these problems??
King
  • King
is the number 202,or 404 or 101 counted ?
Mr.Math
  • Mr.Math
I would add to Ishaan solution that \[100A+10B+C\ge 100C+10B+A \implies A\ge C.\]
Mr.Math
  • Mr.Math
This is probably the only missing piece in his solution.
King
  • King
guys are palindromic nos. counted?
King
  • King
palindromic even nos.?
anonymous
  • anonymous
Hmm according to me I need even numbers for A and C of the form ABC. \[A \in {2,4,6,8}\] Oh yeah, Mr.Math is right.
Mr.Math
  • Mr.Math
Oh or \(A>C\), since you said positive (i.e. not including difference of 0).
King
  • King
so, ishaan is 202,404,606,808 rite?
anonymous
  • anonymous
Then the number is certainly less than 200. No@ King, the question says positive multiples. The number must be for 200s 0. For 400s 9 (412, 422, 432, ...). For 600s 18 (I think not sure) and finally for 800s 27. Is it 9 + 18 + 27 = 9(1 + 2+3) = 54?
anonymous
  • anonymous
The answer is 60.
anonymous
  • anonymous
Oh no :-/ I must have done something silly.
King
  • King
there are 21 in 600's and 30 in 800's so answer is 60
anonymous
  • anonymous
Oh yeah, How can I forget 802, 804, 806, 604, 602 and 402!?!
anonymous
  • anonymous
Well done ishaan :)
anonymous
  • anonymous
Thanks! Took me 40 minutes to solve a very very easy problems of yours, I wonder how much hard one would take.
anonymous
  • anonymous
hehe :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.