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anonymous
 4 years ago
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!
anonymous
 4 years ago
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!

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no i found one for the 1st one its not 0....wait...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am sorry Ishaan but that's not the right answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{{AA*100} + BA}{3*11} = \frac{AA*100}{11*3} + \frac{BA}{11*3}\] 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
 4 years ago
Best ResponseYou've already chosen the best response.0\( 3333 \) is a valid isn't? ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But you said AABA? Hmm if B=A is valid then, {3333, 6666, 9999} are clear solutions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I never said \(B \neq A \) ;)

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.01) All numbers in the form \(AABA\) such that: i) \(3A+B=k\), where k is a number divisible by 3, and ii) \(BA=c\), 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 \[BA=0 \implies B=A\] and \[3A+B=0, 3, 6,.. 36\] From here we can conclude that the numbers are {0000, 3333, 6666, 9999}.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.00000 is 0 not a 4 digit no. ...........

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm 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(AC) = 2*33*k \[\frac{3}{2} * (AC) = k\] 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
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm, I need to check the definition of a digit. But the idea is clear, I think.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ishaan dont we have to find the amnt?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.00000 is not a four digit number.

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0Then 3 such numbers exist.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Try the second one now :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i fnd 1 number fr the 2nd one..... 400

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0I think Ishaan already did it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but then we hav o find the amnt

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0I would rather let him find it, or I would be stealing his work :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ishaan's answer is probably 4*10*5 but that doesn't seem to be the correct answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, I wasn't trying to find the number, Limit of laziness. Let me try now!

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0How do you learn how to solve these problems??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is the number 202,or 404 or 101 counted ?

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0I would add to Ishaan solution that \[100A+10B+C\ge 100C+10B+A \implies A\ge C.\]

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0This is probably the only missing piece in his solution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0guys are palindromic nos. counted?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0palindromic even nos.?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm 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
 4 years ago
Best ResponseYou've already chosen the best response.0Oh or \(A>C\), since you said positive (i.e. not including difference of 0).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so, ishaan is 202,404,606,808 rite?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then 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
 4 years ago
Best ResponseYou've already chosen the best response.0Oh no :/ I must have done something silly.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there are 21 in 600's and 30 in 800's so answer is 60

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh yeah, How can I forget 802, 804, 806, 604, 602 and 402!?!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks! Took me 40 minutes to solve a very very easy problems of yours, I wonder how much hard one would take.
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