In the second subtraction, the number being subtracted is three digits, and the first digit isn't 1. So it must be at least 200, and the divisor is at least 21. Since the first digit of the divisor isn't 2, the divisor must be at least 30.
In the first subtraction, the number being subtracted is at least twice the divisor, but is only two digits. So the ...
You are looking for least percentage, without concerning with the Difficulty of the puzzle.
In that case, the percentage can be made arbitrarily small.
Here is an example with 31 digits removed:
Here the format AAA)BBB(CCC means AAA is dividing BBB to get the ...
I think I found the answer.
Step one to find the divisor.
I started with
Now this number start making sense in my mind. Here's how
Second step of division.
Third step of division.
Fourth/Final step of division
Here is my actual work
_,___ | ( 1000)
x _,___ | ( 1000)
_,___,___ | (1000000)
1,___,___ | (1000000)
is not uniquely solvable; how about e.g.
On the other hand, isn't the following uniquely solvable?
First I will replace the $X$'s with different lowercase letters, so its easier to refer to them.
F a 8 b c
A d C | e f B D g h i j
k m J
G n g h
o p H
q r i j
s t u E
I ran a computer program and I was wondering why it didn't find a solution. It's because some numbers presented as 4-digit number are in fact 3-digit numbers.
So I ran it again and it found 2350 solutions!
20 solutions have XXBDXXXX mod AXC = 0. (The problem didn't say it had to be 0)
Here's all 20 solutions with mod 0:
124 * 80800 = 10019200 (A,B,C,D,E,F,...
As a way of reiterating @VassilisParassidis' answer from another perspective, the puzzle is reminiscent of a "connect wall" puzzle of which there are many examples on puzzle SE (e.g. here, here, and here.)
The thing to note is that the 4x4 layout is not ordered in any particular way - the items could just as easily be given as an unordered list of ...
It's probably a typo
Given that there are only 2 examples to work with, it must be a very simple formula. Usually the formula uses each number once.
I wrote a computer search for all ways to combine the 4 numbers around the circle with operations +, -, *, / in a way that it produces the value in the circle for the left and right case. The only formulas ...
The missing number is 6.
Inside the square we have 2,7-3,4-2,6-1,1-1,8-1,9-2,5-1,2-2,3. If we use all numbers taken four at a time regardless of combinations the missing number is always 6 when the sum of the four numbers is equal to 20.
More logical explanatios exist, but I prefer the one I gave above.
Ok so this has puzzled me for a while. I have a partial answer based on some assumptions. This test is designed for primary school children, looking at the rest of the paper everything is basic so this puzzle should be the same. I think we are supposed to ignore the bottom horizontal line and instead picture the shape in 8 equal slices. I think the two ...