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anonymous

  • 5 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!

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  1. anonymous
    • 5 years ago
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    1. 0?

  2. King
    • 5 years ago
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    no i found one for the 1st one its not 0....wait...

  3. anonymous
    • 5 years ago
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    I am sorry Ishaan but that's not the right answer.

  4. anonymous
    • 5 years ago
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    \[\frac{{AA*100} + BA}{3*11} = \frac{AA*100}{11*3} + \frac{BA}{11*3}\] 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?

  5. anonymous
    • 5 years ago
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    \( 3333 \) is a valid isn't? ;)

  6. anonymous
    • 5 years ago
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    But you said AABA? Hmm if B=A is valid then, {3333, 6666, 9999} are clear solutions.

  7. anonymous
    • 5 years ago
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    I never said \(B \neq A \) ;)

  8. Mr.Math
    • 5 years ago
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    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}.

  9. King
    • 5 years ago
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    answer is 3?

  10. Mr.Math
    • 5 years ago
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    4.

  11. King
    • 5 years ago
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    0000 is 0 not a 4 digit no. ...........

  12. anonymous
    • 5 years ago
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    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.

  13. Mr.Math
    • 5 years ago
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    Hmm, I need to check the definition of a digit. But the idea is clear, I think.

  14. King
    • 5 years ago
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    @ishaan dont we have to find the amnt?

  15. anonymous
    • 5 years ago
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    0000 is not a four digit number.

  16. King
    • 5 years ago
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    yes!!!

  17. Mr.Math
    • 5 years ago
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    Then 3 such numbers exist.

  18. anonymous
    • 5 years ago
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    Try the second one now :)

  19. King
    • 5 years ago
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    i fnd 1 number fr the 2nd one..... 400

  20. Mr.Math
    • 5 years ago
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    I think Ishaan already did it.

  21. King
    • 5 years ago
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    but then we hav o find the amnt

  22. King
    • 5 years ago
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    is 0 div by 66?

  23. Mr.Math
    • 5 years ago
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    I would rather let him find it, or I would be stealing his work :P

  24. anonymous
    • 5 years ago
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    Ishaan's answer is probably 4*10*5 but that doesn't seem to be the correct answer.

  25. King
    • 5 years ago
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    is 0 div. by 66.....

  26. anonymous
    • 5 years ago
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    Oh, I wasn't trying to find the number, Limit of laziness. Let me try now!

  27. NotTim
    • 5 years ago
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    How do you learn how to solve these problems??

  28. King
    • 5 years ago
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    is the number 202,or 404 or 101 counted ?

  29. Mr.Math
    • 5 years ago
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    I would add to Ishaan solution that \[100A+10B+C\ge 100C+10B+A \implies A\ge C.\]

  30. Mr.Math
    • 5 years ago
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    This is probably the only missing piece in his solution.

  31. King
    • 5 years ago
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    guys are palindromic nos. counted?

  32. King
    • 5 years ago
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    palindromic even nos.?

  33. anonymous
    • 5 years ago
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    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.

  34. Mr.Math
    • 5 years ago
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    Oh or \(A>C\), since you said positive (i.e. not including difference of 0).

  35. King
    • 5 years ago
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    so, ishaan is 202,404,606,808 rite?

  36. anonymous
    • 5 years ago
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    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?

  37. anonymous
    • 5 years ago
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    The answer is 60.

  38. anonymous
    • 5 years ago
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    Oh no :-/ I must have done something silly.

  39. King
    • 5 years ago
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    there are 21 in 600's and 30 in 800's so answer is 60

  40. anonymous
    • 5 years ago
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    Oh yeah, How can I forget 802, 804, 806, 604, 602 and 402!?!

  41. anonymous
    • 5 years ago
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    Well done ishaan :)

  42. anonymous
    • 5 years ago
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    Thanks! Took me 40 minutes to solve a very very easy problems of yours, I wonder how much hard one would take.

  43. anonymous
    • 5 years ago
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    hehe :)

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