Question

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!

Answer 1

1. 0?

Answer 2

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

Answer 3

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

Answer 4

\[\frac{{AA*100} + BA}{3*11} = \frac{AA*100}{11*3} + \frac{BA}{11*3}\] 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?

Answer 5

\( 3333 \) is a valid isn't? ;)

Answer 6

But you said AABA?

Hmm if B=A is valid then, {3333, 6666, 9999} are clear solutions.

Answer 7

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

Answer 8

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}.

Answer 9

answer is 3?

Answer 10

4.

Answer 11

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

Answer 12

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.

Answer 13

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

Answer 14

@ishaan dont we have to find the amnt?

Answer 15

0000 is not a four digit number.

Answer 16

yes!!!

Answer 17

Then 3 such numbers exist.

Answer 18

Try the second one now :)

Answer 19

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

Answer 20

I think Ishaan already did it.

Answer 21

but then we hav o find the amnt

Answer 22

is 0 div by 66?

Answer 23

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

Answer 24

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

Answer 25

is 0 div. by 66.....

Answer 26

Oh, I wasn't trying to find the number, Limit of laziness.

Let me try now!

Answer 27

How do you learn how to solve these problems??

Answer 28

is the number 202,or 404 or 101 counted ?

Answer 29

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

Answer 30

This is probably the only missing piece in his solution.

Answer 31

guys are palindromic nos. counted?

Answer 32

palindromic even nos.?

Answer 33

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.

Answer 34

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

Answer 35

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

Answer 36

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?

Answer 37

The answer is 60.

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

Well done ishaan :)

Answer 42

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

Answer 43

hehe :)