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More ways to get Braingle...

The Winning Combination 2

Logic puzzles require you to think. You will have to be logical in your reasoning.

 

Puzzle ID:#34548
Fun:*** (2.78)
Difficulty:*** (2.5)
Category:Logic
Submitted By:grungy49Aau******
Corrected By:MarcM1098

 

 

 



You are back on the same game show with four other contestants. You didn't realise last time that the winner gets another appearance on the show! You now have another chance to win half a million dollars and become a millionaire! The objective is to crack the combination of the safe using the clues. The first person to do so will win $500,000.

The safe combination looks like this:

??-??-??-??

A digit can be used more than once in the code, and there are no zeroes.

Here are the clues:

1. When adding the first number of the third set with the second number of the second set, you will get a number below 8.

2. The only digits not used in the code are 5 and 7.

3. When adding the two digits of the first set together, you will not get 10.

4. If you multiply the second set by 4, you will get the fourth set reversed.

5. No set has a number above 90.

6. The product of both the digits in the first set is below 30 but above 20.

7. One of the number sets is 66.

8. There is only one 8 in the combination, and it is in the last set.

Now, get cracking!

Warning: Hint shows which set the 66 is in.



Answer

The set is 39-21-66-48.

First of all, the second clue tells you there are no 5's or 7's in the combination.

Now we go to the fourth clue. It tells you that if you multiply the second set by 4, you will get the fourth set reversed. There are only a few numbers that will satisfy this:

11, 12, 16, 21, 22, 23 and 24.

Now we skip down to clue 8, which tells us there is one 8 only, and it is in the last set. It could either be the first digit or the second digit.

However, this eliminates these numbers from being in the second set: 11, 16, 23 and 24. Now, only 12, 21 and 22 are the possibilities for being in the second set.

Now, let's look at clue 7. This tells us that 66 is one of the four number sets. We know it can't be in the second, or fourth, and since clue six tells us the product of the two numbers in the first set ranges from 21-29, 66 must go in the third set.

??-??-66-??

Now we must look at the first clue. The first number of the third set (6) plus the second number of the second set will get a number below 8. This means it cannot be 12 or 22, because they would both total 8. This means that the second set of numbers is 21. Making good progress!

??-21-66-??

Now, we come back to clue 4 which tells us the second set multiplied by 4 equals the fourth set reversed. 21 multiplied by 4 is 84, which when reversed is 48!

??-21-66-48

Now, we come back to clue 6, which tells us the product of the digits in the first set is above 20 but below 30. These are the only numbers that satisfy that condition - and remember, no 5's, 7's or 8's are left in the combination:

39 (3 * 9 = 27)
46 (4 * 6 = 24)
64 (6 * 4 = 24)
93 (9 * 3 = 27)

Now, we look at clue 5. This eliminates 93. Now, come to clue 3. This tells us the sum of the two digits in the first set does not equal 10. This eliminates 46 and 64. The only number left is.... 39!

Now you have your answer;

39-21-66-48

Congratulations!!! You won, again!! You are over the moon! You have become the first millionaire in five generations of your family history!!

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