I get 35.
Shapes are worth 5 each (15 for all 3)
Bananas are 1 each (4 for a group of 4)
Clock is 1 per hour on clock (3 with 3 o'clock showing)
So:
(2 o'clock) + (3 bananas) + [(3 bananas) x (2 shapes)]
= 2 + 3 + [3 x 10]
= 2 + 3 + 30
= 35
I get 35.
Shapes are worth 5 each (15 for all 3)
Bananas are 1 each (4 for a group of 4)
Clock is 1 per hour on clock (3 with 3 o'clock showing)
So:
(2 o'clock) + (3 bananas) + [(3 bananas) x (2 shapes)]
= 2 + 3 + [3 x 10]
= 2 + 3 + 30
= 35
Count the sides of the geometrics, not the quantity. 38.
Fields the transaction did not carry are omitted. Open the payload to see the bytes as stored.
1EZKvwVyKb9LtwnHCtXAneub8w5WuoX3oL VerifiedAh! I figured I was missing something with the shapes, but I couldn't put my finger on it...!
It's no coincidence the triple shape adds up to 15 both ways I'm sure, but 'each shape is worth 5' doesn't fit the rest of this puzzle's logic.
Agreed.
By this interpretation it’s indeterminate.
🖼 vs 🖼
Not at all. The bananas define it. Four bananas is 4, three is 3. For it to be indeterminate, any quantity of bananas would have to be equivalent to any other quantity, the same way Peter thought a 4-sided geometric was equivalent to a 5-sided one.