## 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!

1. anonymous

1. 0?

2. anonymous

no i found one for the 1st one its not 0....wait...

3. anonymous

I am sorry Ishaan but that's not the right answer.

4. 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?

5. anonymous

$$3333$$ is a valid isn't? ;)

6. anonymous

But you said AABA? Hmm if B=A is valid then, {3333, 6666, 9999} are clear solutions.

7. anonymous

I never said $$B \neq A$$ ;)

8. 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}.

9. anonymous

10. Mr.Math

4.

11. anonymous

0000 is 0 not a 4 digit no. ...........

12. 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.

13. Mr.Math

Hmm, I need to check the definition of a digit. But the idea is clear, I think.

14. anonymous

@ishaan dont we have to find the amnt?

15. anonymous

0000 is not a four digit number.

16. anonymous

yes!!!

17. Mr.Math

Then 3 such numbers exist.

18. anonymous

Try the second one now :)

19. anonymous

i fnd 1 number fr the 2nd one..... 400

20. Mr.Math

I think Ishaan already did it.

21. anonymous

but then we hav o find the amnt

22. anonymous

is 0 div by 66?

23. Mr.Math

I would rather let him find it, or I would be stealing his work :P

24. anonymous

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

25. anonymous

is 0 div. by 66.....

26. anonymous

Oh, I wasn't trying to find the number, Limit of laziness. Let me try now!

27. NotTim

How do you learn how to solve these problems??

28. anonymous

is the number 202,or 404 or 101 counted ?

29. Mr.Math

I would add to Ishaan solution that $100A+10B+C\ge 100C+10B+A \implies A\ge C.$

30. Mr.Math

This is probably the only missing piece in his solution.

31. anonymous

guys are palindromic nos. counted?

32. anonymous

palindromic even nos.?

33. 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.

34. Mr.Math

Oh or $$A>C$$, since you said positive (i.e. not including difference of 0).

35. anonymous

so, ishaan is 202,404,606,808 rite?

36. 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?

37. anonymous

38. anonymous

Oh no :-/ I must have done something silly.

39. anonymous

there are 21 in 600's and 30 in 800's so answer is 60

40. anonymous

Oh yeah, How can I forget 802, 804, 806, 604, 602 and 402!?!

41. anonymous

Well done ishaan :)

42. anonymous

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

43. anonymous

hehe :)