(Puzzles) Challenging Mathematical Puzzles (20)

Puzzles : Challenging Mathematical Puzzles

57. There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?

Answer : -
It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.

1. Take 8 coins and weigh 4 against 4.
* If both are not equal, goto step 2
* If both are equal, goto step 3


2. One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.
* If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.
    o If both are equal, L4 is the odd coin and is lighter.
    o If L2 is light, L2 is the odd coin and is lighter.
    o If L3 is light, L3 is the odd coin and is lighter.


* If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2
    o If both are equal, there is some error.
    o If H1 is heavy, H1 is the odd coin and is heavier.
    o If H2 is heavy, H2 is the odd coin and is heavier.


* If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4
    o If both are equal, L1 is the odd coin and is lighter.
    o If H3 is heavy, H3 is the odd coin and is heavier.
    o If H4 is heavy, H4 is the odd coin and is heavier.


3. The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.
    * If both are equal, there is some error.
    * If X is heavy, X is the odd coin and is heavier.
    * If X is light, X is the odd coin and is lighter.

 

58. In a sports contest there were m medals awarded on n successive days (n > 1).
1. On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.
2. On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.
3. On the nth and last day, the remaining n medals were awarded.

How many days did the contest last, and how many medals were awarded altogether?

Answer :- 


Total 36 medals were awarded and the contest was for 6 days.

On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals
On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals
On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals
On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals
On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals
On day 6: Medals awarded 6

I got this answer by writing small program. If anyone know any other simpler method, do submit it.

 

59. A number of 9 digits has the following properties:

- The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9.
-  Each digit in the number is different i.e. no digits are repeated.
-  The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.

Find the number.

Answer :- 

The answer is 381654729

One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur. 

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