There are 16 possible solutions: Assuming a x b = c I + j = k x - y = z
You start of by finding that c can only equal 6 or 8 meaning a or b must equal 2. Since you can also conclude that either k or x must equal 9 you can arrive with 12 different equations. When k or x = 9 1,8 3,6 4,5 5,4 6,3 8,1
We will ignore the last there as they are just the other way around. We can find out that there is no way to use 1,8 or 3,6 but is a way for 4,5. We also know that a can switch with b, I can switch with j and y can switch with z. This gives 8 possible rearrangements.
Since x or k could equal 9 there is double 8 amount of solutions making it 16.
The square ABCD is a unit square. Next to it is square BEFG. I am interested in the triangle AFC. What is its area? Why is this a surprising question? What happens if square BEFG is smaller than ABCD? There is an interactive version of this image at https://ggbm.at/BphKr2mG
Binary is the name given to numbers in base 2. This means the only digits that can be used are 0 and 1. We use the powers of 2 as column headings. Here is an example: This shows 22 in binary, because 16 + 4 + 2 = 22 How can we write 43 in binary? Try it first before scrolling down (but it is then worth doing so, because there are at least two different ways of doing this!). . . . . . . . . We could start at the left-hand end and see whether we need each number. So: the biggest power of 2 that fits in 43 is 32. (We don’t need to worry about 64, 128, 256, etc.) Put a 1 in the 32 column and subtract 32 from 43, leaving 11. There are no 16s in 11, so put a zero in the 16 column. There is an 8, so put a 1 in the 8 column and subtract that from 11 to leave 3. So far this is the state of play: We have 3 left. This is no 4s, one 2 and one 1. So 43 = 101011 in b...
There are 16 possible solutions:
ReplyDeleteAssuming
a x b = c
I + j = k
x - y = z
You start of by finding that c can only equal 6 or 8 meaning a or b must equal 2.
Since you can also conclude that either k or x must equal 9 you can arrive with 12 different equations.
When k or x = 9
1,8
3,6
4,5
5,4
6,3
8,1
We will ignore the last there as they are just the other way around.
We can find out that there is no way to use 1,8 or 3,6 but is a way for 4,5.
We also know that a can switch with b, I can switch with j and y can switch with z. This gives 8 possible rearrangements.
Since x or k could equal 9 there is double 8 amount of solutions making it 16.