Question

A king asks a learned man to visit his palace. The man says I will come any day next month. But give me gold in gms equal to the date on which I come. The king asks the jeweler to make rings of 1 to 31 gms. The jeweler was smart and makes only 5 rings. What was the weights of those rings?

Answer 1

We be solving the problem usig two different method:

Method (1) : Trial and error ( If we are not aware of any formula)

Day 1   :    1gm (New ring)----------------------------------------(Ring 1)
Day 2   :    2gm (New ring)----------------------------------------(Ring 2)
Day 3   :    3gm ( combining 1gm  and 2gm)
Day 4   :    4gm ( New ring)----------------------------------------(Ring 3)
Day 5   :    5gm ( combining 1gm  and 4gm)
Day 6   :    6gm ( combining 2gm  and 4gm)
Day 7   :    7gm ( combining 1gm, 2gm  and 4gm)
Day 8   :    8gm ( New ring)-----------------------------------------(Ring 4)
Day 9   :    9gm ( combining 1gm  and 8gm)
Day 10 :  10gm ( combining 2gm  and 8gm)
Day 11 :  11gm ( combining 1gm, 2gm and 8gm)
Day 12 :  12gm ( combining 4gm and 8gm)
Day 13 :  13gm ( combining 1gm, 4gm and 8gm)
Day 14 :  14gm ( combining 2gm, 4gm and 8gm)
Day 15 :  15gm ( combining 1gm, 2gm, 4gm and 8gm)
Day 16 :  16gm ( New ring)---------------------------------------------(Ring 5)
Day 17 :  17gm ( combining 1gm, and 16gm)
Day 18 :  18gm ( combining 2gm and 16gm)
Day 19 :  19gm ( combining 1gm, 2gm and 16gm)
Day 20 :  20gm ( combining 4gm and 16gm)
Day 21 :  21gm ( combining 1gm, 4gm and 16gm)
Day 22 :  22gm ( combining 2gm, 4gm and 16gm)
Day 23 :  23gm ( combining 1gm, 2gm, 4gm and 16gm)
Day 24 :  24gm ( combining 8gm and 16gm)
Day 25 :  25gm ( combining 1gm, 8gm and 16gm)
Day 26 :  26gm ( combining 2gm, 8gm and 16gm)
Day 27 :  27gm ( combining 1gm, 2gm, 8gm and 16gm)
Day 28 :  28gm ( combining 4gm, 8gm and 16gm)
Day 29 :  29gm ( combining 1gm, 4gm, 8gm and 16gm)
Day 30 :  30gm ( combining 2gm, 4gm, 8gm and 16gm)
Day 31 :  31gm ( combining 1gm, 2gm, 4gm, 8gm and 16gm)

Therefore, the five rings are - 1gm, 2gm, 4gm, 8gm, and 16gm.

Method 2 : Theorem- A natural number greater than 1 is a sum of consecutive integers if and only if it is not a power of 2.
(Observe the answer, all are consecutive powers of 2 -
$1=2^0 \\ 2=2^1 \\ 4=2^2 \\ 8=2^3 \\ 16=2^4$
)
Concept we will be using here is " n consecutive powers of two will give all the whole numbers   till $2^{(n+1)}-1$

We know that $2^5=32$ but the maximum weight is 31gm.
Then, maximum power of 2 will be 4.

Then, all the powers of 2 will be 4, 3, 2, 1, 0
Therefore, the rings are -
$2^4=16 gm \\ 2^3=8gm \\ 2^2 =4gm \\ 2^1=2 gm \\ 2^0=1gm$



Answer:  1gm, 2gm, 4gm, 8gm, and 16gm.
 

Answer 2

He made five rings of weights 1gm 2gm 4gm 8gm and 16gm by combining these 1gm to 31gm can be made. Because they cover all grams required by the king.