1. WordPlay I am a word with six letters. When one letter is removed, I am now 12. What word am I? (Sent along by subscriber Anita Dixon. Her quick view of the original and my poor memory may have altered the translation somewhat.)
2. Digital Dilemma Find a 5-digit number where the sum of all the digits is 25, and each digit is 2 more than the digit preceeding it.
3. HUH? 🙂 What number is 20 more than twice the number that is 10 more than 15 times one-half of ten?
4. Counterfeit Coin, Part 1 You have 12 coins in front of you, knowing that one of them is counterfeit. All you know is that the ‘bad’ coin is slightly heavier than the others. You also have a balance scale in front of you, and your task is to determine the ‘bad’ coin using only three (3) weighings*. How do you accomplish this?
*A weighing occurs any time you set something on both sides of the scale.
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1. Word with 6 letters = DOZENS
Remove the “S” and you have DOZEN.
2. 13,579
3. I’ll have to work on that one when I have more time. We’re in St. Louis just now, helping Josh with the grandkids.
4. I think there’s more than one solution here, but one of them would be Step 1: put 3 coins on each side . . . throw out the lighter set of three. Step 2: From the heavy set of 3, put one on each scale. If they balance, you know the 3rd one is the heavy one. If they don’t balance, you know the heavier one is the bad coin.
If you start by putting 2 on each scale, you can accomplish the feat in 2 steps.
#1: dozens less the s becomes dozen
#2: 1,3,5,7,9
#3: 190
#4: Weigh all coins with six on each side of scale. Take the group of six that weighs heaviest, and weigh three to a side. Keep the group of three that weighs heaviest. Weight any two coins. If one weighs heavier, that is the “bad” coin. If the two coins weigh the same, the third coin (the one not weighed) is the “bad” coin.
#3. I think that number is 190.
Attempting 1) and 2)! Ignoring 3 and 4!
1) dozens. Remove the s and it is dozen…12.
2) 13,579