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A possible 3rd solution! If not. I'm giving up!

Two solutions. Maybe more?

It certainly should. It clearly fails a units check. Good catch.

The subtlety is…

Al, Bert, and Charlie competed in a track and field event in which points were awarded for 1st, 2nd, and 3rd, place only. At the end of the day, Al had accumulated 22 points, while Bert and Charlie each garnered only 9 points. No other competitor earned points. Bert was 1st in the shot put. Who finished 2nd in the javelin throw? This is a Gold star puzzle.

answer

EDIT: For clarity, I use "number" to reference an individual digit. I use "sequence" to reference the string of numbers. EDIT2: Added spoiler tag.

Perhaps your question may hinge on the source of variability. Pistol has 5-7. In any skirmish, is the choice between 5, 6, 7 made by the program randomly? Or does it depend upon distance to target? Or visibility ( like heavy, light or no fog)? And, as Pickett points out, damage radius is a key parameter of the Dupuy lethality index.

Another factor to possibly consider...

Depends...

Fuzzy thoughts:

You are shown a pile of dimes all of which have one of two distinct weights differing by a small amount not detectable by feel. Forty eight dimes are separated from this pile and you are told of these forty eight, light ones are a dime a dozen (literally - i.e. 44 heavy dimes and 4 light dimes). Using a balance scale twice, find seven heavy dimes. EDIT: for clarification

4 light ones/48

Ten years ago I called attention to a number that when divided by a single integer p it left a remainder of p-1. (Help, a remainder is chasing me) Here is a chance to construct a nine-digit number, a permutation of { 1 2 3 4 5 6 7 8 9 } that has no remainders, sort of. The task is to permute { 1 2 3 4 5 6 7 8 9 } to create a number whose first n digits is a multiple of n for any single-digit n. For example, consider 123654987. Its first 2 digits (12) are divisible by 2. It's first 5 digits (12365) are divisible by 5. However this is not a solution, since 1236549 is not a multiple of 7.

Conjoined twins?

@flamebirde because Molly was playing off of the word easy, which is a popular Commodores song. No other reason

Doesn't look like people get many responses in this section but I will introduce myself anyway. I'm a 36-year-old mom. I used to be smart, a long time ago. Was in the gifted program throughout grade school. Battled severe psychological problems starting in high school and am still battling them today. My grades in high school and college were pretty bad and I dropped out of pre-med. My younger sisters (who were not "gifted") have achieved much more academically and professionally than I have. Which is fine, because I like my life. But, sometimes I miss people telling me I'm smart! SO, I have started playing around with Mensa-type puzzles in an attempt to stretch my brain. I'm not having a lot of luck so far-- which is why I'm here.

Agree. When I hit the send button, I realized my thinking was too simple. But instead of deleting my post (moderator privilege) I left it to take its licks.

I believe you replace the ? with digits.

I swear: 1) To strangle the next person who uses 'suicide' as a verb. 2) That if I offended or hurt you in any way, I didn't mean it. 3) That I'll stop procrastinating. Tomorrow. Add whatever you swear.

Excellent riddle Plasmid. Those types are my favorite. Give yourself a 5 star vote and make it official. BTW . . . what's the "vote" thing for?

Depends on what is meant by "have." [1] If have means "own," then yes. I can give someone something that I do not own. e.g. if I stole it. To give something, one only needs the ability to determine who controls it. If I control it, I can pass its control to someone else. So ... [2] If have means "possess the control of" then no. As stated, the paradox arises from the different antecedents of "with sorrow." Sorrow is the consequence of giving, not a possession before the act. But the language permits that interpretation by its form. Cute.