two wands in the last pic also. ;)

There's a second layer of complexity in it, which is pretty obvious when you look at it closely. It catches people not paying close attention. I credit 71T Targa for busting it wide open.

I go with RB except using his reasoning I come up with 17.

3 wands value each = 7
4 brooms value each = 3
Bottom row: 2 wands + 1 broom.

I can't see that, but I'm old and the picture is intentionally fuzzified.

How many brooms?

4. 2 and a double image.

This is correct..

The Question is http://forums.pelicanparts.com/uploa...1586332298.jpg

although there are times when "order of addition" is critical. Ever made emulsions?

This one is more challenging, but not too difficult:

http://forums.pelicanparts.com/uploa...1586372331.jpg

Yep..lots of them in college. Liquor was cheap then. :)

012

You broke the second rule.

102

Breaks rule #3.

Has to be: 042

On the original question:

All 3 witches,X on the top row are carrying a star wand,Y and a broom,Z.
3X+3Y+3Z = 45
(assume addition, instead of a multiplier or a different symbol)

All 3 star wands are the same.
3Y=21
Star wand =7

There are actually 4 brooms in the third row. I didn't notice at first.
(assume addition, instead of a multiplier or a different symbol)
4Z=12
Broom= 3

Going back to the first line to find value of the witch, X:
45 - three wands(=21) - three brooms(=9) = 15. 15 divided by three witches is 5.
Witch = 5.

So 7 + 3 + 5 =15

042

Edit nm already answered.

Your numbers for the individual items is correct. However, there are 2 wands in final equation, and the later action is x not +.

So considering PEMDAS it's: 3+(5x14)=73

(Steve nailed in on post #11).

;)

I first got 102 until I re-read #3.
Eric is right. Again..

Good catch.