The Spotting Contest

During a recent plane and train spotting contest, five eager entrants were lined up ready to be tested on their spotting ability. 

They had each spotted a number of planes (26, 86, 123, 174, 250) and a number of trains (5, 42, 45, 98, 105). From the clues below, can you determine what colour shirt each was wearing, their position, their age (21, 23, 31, 36, 40) and the number of trains and planes spotted?

1. Bob spotted 44 less trains than planes. 

2. Tom was 36 years old. 

3. The person on the far right was 8 years younger than Bob, and spotted 174 planes. 

4. Josh was wearing a yellow shirt and spotted 37 trains fewer than Bob. 

5. The person who was wearing a green shirt, was 19 years younger than the person to his left. 

6. Steven spotted 105 trains and 250 planes.

7. The person in the center was 31 years old, was wearing a blue shirt and spotted 42 trains.

8. Doug, who was on the far left, spotted 26 planes, and spotted 72 trains more than planes. 

9. The person who was wearing a red shirt was 4 years older than Tom and was not next to the person wearing a blue shirt. 

10.The person who was next to the 31 year old, but not next to the person who spotted 26 planes, was wearing a orange shirt, and spotted 45 trains. 

Here are final STATS of the contest! 

Final Stats of The Spotting Contest

What was the contest?

Five participants having ages (21, 23, 31, 36, 40) had each spotted a number of planes (26, 86, 123, 174, 250) and a number of trains (5, 42, 45, 98, 105). 

Below are the clues given - 


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1. Bob spotted 44 less trains than planes. 

2. Tom was 36 years old. 

3. The person on the far right was 8 years younger than Bob, and spotted 174 planes. 

4. Josh was wearing a yellow shirt and spotted 37 trains fewer than Bob. 

5. The person who was wearing a green shirt, was 19 years younger than the person to his left. 

6. Steven spotted 105 trains and 250 planes. 

7. The person in the center was 31 years old, was wearing a blue shirt and spotted 42 trains. 

8. Doug, who was on the far left, spotted 26 planes, and spotted 72 trains more than planes. 

9. The person who was wearing a red shirt was 4 years older than Tom and was not next to the person wearing a blue shirt. 

10. The person who was next to the 31 year old, but not next to the person who spotted 26 planes, was wearing a orange shirt, and spotted 45 trains. 

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STEPS : 

Let's make at table like below and fill the data one by one.

 
STEP 1 : 

As per (3), the person on the far right that is at position 5 had spotted 174 planes & he must be 23 years old so that Bob aged 31 years (only pair of ages having 8 years difference.

 
STEP 2 : 

As per (7), the person at position 3 who is wearing blue shirt and is 31 year old. He has spotted 42 trains.

STEP 3 : 

For (5) to be true, the ages of those participants occupying consecutive positions must be 40 and 21 respectively. Only, positions lefts for (5) to be true are 1 & 2.

STEP 4 : 

And as per (8), the person at position 1 must be Dough who spotted 26 planes and spotted 72  trains more than planes i.e. 26 + 72 = 98 trains.

STEP 5 : 

As per (10), the person at position 4 must be wearing orange shirt and must have spotted 45 trains. His age must be 36 years (only number in ages list left).

STEP 6 :

As per (9) suggests, Tom must be 36 years old positioned at 4 and Dough must be the person wearing red shirt. With that, Josh in (4) must be wearing yellow shirt and positioned at no.5. 

STEP 7 : 

The hint (4), also suggests that Josh must had spotted 5 trains out of 42-5 possible spotted trains pair shared with Bob. Hence, the person at 2nd position must had spotted 105 trains (only number left in list of number of trains spotted).
 

STEP 8 : 

As per (3), the person having age 31 must be Bob. Hence, at no.2, Steven must be there. 

STEP 9 : 

As per (6), Steven have spotted 250 planes. As per (1), Bob must have spotted 42 + 44 = 86 planes. And the only number left as number of planes spotted for Tom is 123.

STEP 10 :

Final Stats look like as - 

Crack And Win $500,000 - Puzzle

You are in a game show with four other contestants. The objective is to crack the combination of the safe using the clues, and 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 leading zeroes.

Here are the clues:

1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.

And get moving, I think another contestant has almost figured it out!

Click here for the SOLUTION! 

Crack And Win $500,000 Puzzle - Solution

What was the puzzle?

We know, the safe has a lock having 4 sets of 2 digits as -

?? - ?? - ?? - ??

Since, leading zeros are not allowed any of the set can't be started with 0 like 01, 07, 09 etc. 

Take a look at the clues given - 

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1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.


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STEPS :

1] As per clue (2), digits 2, 3, 4 or 5 aren't allowed in any set. That means only digits 0, 1, 6, 7, 8, 9 are allowed for sure.

2] For (3) to be true, possible combinations of 3rd and 4th sets are - 

10 x 4 = 40
11 x 4 = 44
16 x 4 = 64
17 x 4 = 68
18 x 4 = 72
19 x 4 = 76

The third set can't be 10 for 2 reasons. - 1} As per clue (1), the first set will be 01 and leading 0's are not allowed. 2} The 4th set will have digit 4 which is not allowed.

The third set can't start with 2X, 3X, 4X or 5X as those digits aren't allowed. Moreover, it can't be started with 6X, 7X, 8X as in that case the value of the 4th set will exceed it's maximum possible value of 96 (if digit 2 was allowed) or 76.

Out of all above combinations, only 17 x 4 = 68 and 19 x 4 = 76 are valid combination as rest of combinations have digits that aren't allowed.

So, one thing is sure that the first digit of the third set is 1. And hence the first digit of first set also must be 1 as per clue (1).

3] As per clue (5), the second set can't exceed 20. It can't start with 0. Hence, possible values ranges from 10 to 19. That means, the first digit of the second set is also 1.

4] Now as per clue (4), if you add the first number in the first set with the first number in the second set you will get 8. That means the first number of first set is 8 - 1 = 7. 

As of now, the code looks like : 71 - 1? - 1? - ??

5] If 7 is the first digit of first set then as per clue (1), 7 itself is second digit of the third set.

Now, the code looks as : 71 - 1? - 17 - ??

6] As per clue (3) and rightly deduced as a possible combinations for 3rd & 4th set in STEP 2, the 4th set must be 17 x 4 = 68.

With that, code turns into : 71 - 1? - 17 - 68.

7] Finally, as per clue (6), the second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set. This multiplication i.e. ? x 7 must be equal to 71 + 1 = 72. 
Hence, ? = 9.

The final code looks like : 71 - 19 - 17 - 68.