Comments on: Finish this sequence of equalities…

By: 孟海港

孟海港 — Fri, 18 May 2012 05:12:28 +0000

当然是2啦有几个圈就是几
is 2。how many circles ，it is the num。
用不着算法~ 哈哈

By: 孟海港

孟海港 — Fri, 18 May 2012 05:08:00 +0000

当然是2啦有几个圈就是几

By: TB

TB — Sun, 13 May 2012 04:41:23 +0000

I got the answer to be 3, so I would say there is not enough information to come to the author’s conclusion. I used a simple mapping of digits and added them. All the answers are correct.

0, 6, 9 = 1
1, 2, 3, 4, 5, 7 = 0
8 = 2

Therefore, 2581 = 3

By: HJ

HJ — Sun, 06 May 2012 14:27:07 +0000

Crowell- I was fascinated by your answer. How did you get to it?
But then I realized: Why didn’t you just multiply all the values you gave by 10, and then, instead of dividing by 4, divide by 40??
Or, more simply, why did you bother giving 8 a value of 8 and 0 a value of 4, etc. and then having to divide by 4? Just give each number the value of its number of circles, and then you won’t have to divide at all!!

By: Rodrigo

Rodrigo — Wed, 02 May 2012 04:22:12 +0000

“2581=?”
2581=2
8 tiene dos circulos 8=2
2,5 y 1 no tienen circulos por lo tanto
2=0 5=0 1=0 circulos

By: thereFOUR

thereFOUR — Sat, 28 Apr 2012 09:55:19 +0000

I gave up after thinking 2 hours and read the comments. I feel stupid.

By: Christie

Christie — Thu, 19 Apr 2012 04:03:09 +0000

I love how simple this was to solve, once you knew what to look for, and how interesting it was that so many people came up with math solutions to solve the problem doing mathmatics no preschooler could ever understand. How far we have gone from simplicity.

By: irish joy PHD

irish joy PHD — Sat, 07 Apr 2012 14:00:47 +0000

OMG.. this is so easy and youre having difficult times with..i found the answer for only 2 mins, and it counts the number of holes in the numbers…lol.. such a preschoolers mind riddle

By: Tommaso

Tommaso — Fri, 30 Mar 2012 19:41:40 +0000

Sundar, according to your rule I don’t see how 6666=4 :
6 can be divided by 1, 2, 3 and 6
Apply rule 1
So Count is 2
6666 should be 8 and it isn’t…
So, in my opinion, children solution of counting the circles is still the only solution…

By: romadhoni

romadhoni — Thu, 29 Mar 2012 05:27:47 +0000

you calculate it for too long and too complicated.
see all the numbers there must be a hole or circle. Suppose 0 circle 1, 6 circle 1, circle 2 8, 9 circle 1. So if 0000 = 4. If 2411 = 0, because there is no circle on the numbers. So after this you will know the answer to that question.

By: Sundar

Sundar — Wed, 21 Mar 2012 21:01:01 +0000

Hi Friends

1. Take a count of numbers each number is divisible by excluding the number itself and 1
2. For ’0′ , 0 is always divided by any number so its 1

2581 = 2

For

1. 2 can be divided by 1 and 2
But Apply my rule 1
So Count is 0
2. 5 can be divided by 1 and 5
Apply rule 1
So Count is 0
3. 8 can be divided by 1, 2, 4 and 8
Apply rule 1
So count is 2
4. 1 is divisible by 1
Apply rule 1
So count is 0

Add the total count, correct me if I am wrong

- Sundar -

By: Smola

Smola — Tue, 20 Mar 2012 23:12:04 +0000

So, similar problem can be made with horizontal bar count (2457=>4), or not circles, or left curles and horizontal bars but only if the first one is a circle, or…

By: Daniel Lemire

Daniel Lemire — Tue, 20 Mar 2012 12:34:07 +0000

@Andrew

Solving for a polynomial of degree 19 is not the sought-after solution.

By: Andrew

Andrew — Tue, 20 Mar 2012 09:37:27 +0000

Or you could simply use a polynomial that is able to give you the above results… any mathematician worth his salt should be able to get to this answer.

By: johnpaul

johnpaul — Sat, 17 Mar 2012 13:06:18 +0000

Simply count the closed loops…
8=2 closed loops
9=1 closed loop
0=1 closed loop
all other numbers no loop