FB 'Math' problem.

[stevej37]

two wands in the last pic also.

[Steve Carlton]

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.

[Zeke]

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.

[1990C4S]

Originally Posted by stevej37

two wands in the last pic also.

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

[MBAtarga]

Originally Posted by Steve Carlton

Kind of interesting with its subtleties. Looks like:

The wand = 7 each
There's 4 brooms that = 12, so they = 3 each
The witch with the broom and wand = 15
A witch w/o the broom and wand = 5

Bottom line would be 3 + (5 x 14) = 73

How many brooms?

[Zeke]

Originally Posted by MBAtarga

How many brooms?

4. 2 and a double image.

[rattlsnak]

Originally Posted by Steve Carlton

Kind of interesting with its subtleties. Looks like:

The wand = 7 each
There's 4 brooms that = 12, so they = 3 each
The witch with the broom and wand = 15
A witch w/o the broom and wand = 5

Bottom line would be 3 + (5 x 14) = 73

This is correct..

[svandamme]

The Question is

[flatbutt]

Originally Posted by tadd

...and that is why I am a chemist :-). Forgot about that pesky order of operations thing!

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

[Steve Carlton]

This one is more challenging, but not too difficult:

[stevej37]

Originally Posted by flatbutt

Ever made emulsions?

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

[dad911]

Originally Posted by steve carlton

this one is more challenging, but not too difficult:

012

[Steve Carlton]

You broke the second rule.

[gordner]

102

[Eric Coffey]

Originally Posted by gordner

102

Breaks rule #3.

Has to be: 042

[john70t]

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

[Alan A]

042

Edit nm already answered.

[Eric Coffey]

Originally Posted by john70t

So 7 + 3 + 5 =15

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).

[john70t]

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

[Danimal16]

Originally Posted by gtc

If you look closely, there appears to be a second broom behind the middle one, as well as a second wand in the last formula.

Good catch.