Alok Goyal's Puzzles Page

This was a slightly difficult problem for many people, though can be solved with some persistent trial and error. Girish Tutakne, Pradeep Hengavelli, Suman Saraf, Chandu Karan and Pooja Goyal+Arushi Goyal were the ones to send correct answers.

I do not know if there is an elegant solution to this, requires trial and error. The solution is

Puzzle 119 Answer

The only trick one can think of is that the last digit of the two numbers cannot be “2” as the  multiplication will yield an even number as the last digit. Similarly, the last digits cannot be 3,3 or 7,7 or 7,3 as they yield 9, 9 and 1 respectively as the last digit. Therefore the combinations can be 5,5; 7,5 and 3,5. Beginning with these assumptions, one can build upon the different possibilities and rule them out gradually.

Hope you all enjoyed the puzzle!

This was a slightly difficult puzzle, and I got relatively few correct solutions. From Alok’s Mathematical Circles class, I know that several people were able to get the answer, that included Smiti Mittal (had a very innovative and simple answer) and Anisha Goyal. Other people to send the correct answer included Suman Saraf (complete answer), Chandu Karan and Umesh Mayekar. Well done all!

I am reproducing multiple answers, the one from Alok’s class I found to be the most complete is the following:

A,B,C,D,E,F have 15,14,13,10,11,12 respectively

Opposite sitting players total at end of 5 games must be 25 for each pair. So A must have at least 14 (because someone in other pairs must have at least 13). This means D can have at most 11, and he already has 9 in first two rounds. So D must have a zero in some round. Since order of rounds is not important, lets say D has a zero in 3rd round. On last two rounds, D’s possibilities (given he can only score 2 more points) are 00, 01, 11, or 02. We can test each of these in turn.

One from Smiti, which I also liked very much ( though I am not able to conclude from this answer that this is the only possibility):

Smiti had a great solution. Once it is clear that D must have 5 in the second round, she just flipped the scores for 3rd and 4th rounds relative to 1st and 2nd round. This brought everyone to a score of 10. Then simply get A a 5 in the last round

Hope you all enjoyed the puzzle!

Apologies for not posting last weekend, I was traveling on a short vacation!

This was a relatively easy puzzle, many people sent me the correct answer. Folks who sent me the correct answer include Karan Sharma, Suman Saraf, Sanjeev Dugar, Anirudh Baddepudi, Girish Tutakne, Sri Peddu, Sandeep Verma, Alok Kuchlous, Prakhar, Alankrita Agrawal, Praneeth, Chandu Karan, Pratik Poddar and Rahul Garg – Well done all!

The answer is Feb 28, 29, March 1, 2 and 3 – therefore this is possible only in a leap year.

For any five contiguous numbers, the sum is divisible by 5. Since 63 is not, it can only mean that there is a cut off at the end of the month in the sequence of 5 dates. Since all months end between 28 and 31, we can figure out by simple trial and error that there need to be two dates at the end of the month followed by 1, 2 and 3 of the following month. This implies that the two dates need to add up to 57, which again implies that the dates can only be 28th and 29th, and hence Feb 28 and 29th of a leap year.

Hope you all enjoyed the puzzle!

I got many correct answers for this puzzle as well and a very good response. People who sent correct answers included Chandu Karan, Divya, Suman Saraf (The most complete solution) and Pratik Poddar. I am reproducing the solutions from Suman, done both by common logic and later set of all solutions generated from a computer script (I tried to clean them up for duplicates, but too many to do so :-()

Since we know the 4 corners will be counted twice, it means 55 (sum 1..10) + 4 corners should be a multiple of 4 since all sides all add up to the same sum. The sum of the corners could range between 10 (1,2,3,4 in the corners) to 34 (7,8,9,10) in the corners. This leaves us with the following options:

sum of sides | sum of corners

17 (*4 = 68) | 13
18 (*4 = 72) | 17
19 (*4 = 76) | 21
20 (*4 = 80) | 25
21 (*4 = 84) | 29
22 (*4 = 88) | 33

Using this option we can try and fill the frame. The last one is the simplest because there is only one combination that could fill the corners (i.e. sum of corners is 33) – 6,8,9,10. Using this we could fill it up in the following way:

6 7 9
5 3
1 2
10 4 8

Obviously a few transforms of this arrangement will also suffice.

Similarly, one can find solutions for sums = 18, 19 and 20

1 9 8
4 3
7 5
6 10 2

1 8 10
5 2
6 4
7 9 3

2 8 10
3 1
6 5
9 7 4

Exhaustive list of solutions:

results in:

sum(19)
1 8 10
5 2
6 4
7 9 3

sum(19)
1 8 10
5 2
7 3
6 9 4

sum(18)
1 9 8
4 3
7 5
6 10 2

sum(19)
1 10 8
4 2
5 6
9 7 3

sum(20)
2 8 10
3 1
6 5
9 7 4

sum(18)
2 10 6
3 1
8 7
5 9 4

sum(19)
3 9 7
1 2
10 4
5 8 6

sum(20)
3 9 8
2 1
5 7
10 6 4

sum(22)
6 7 9
1 2
5 3
10 4 8

sum(20)
8 9 3
1 2
7 5
4 6 10

sum(20)
8 9 3
1 5
7 2
4 6 10

sum(20)
8 9 3
7 2
1 5
4 6 10

sum(20)
8 9 3
7 5
1 2
4 6 10

sum(19)
8 10 1
2 4
6 5
3 7 9

sum(19)
8 10 1
2 5
6 4
3 7 9

sum(19)
8 10 1
6 4
2 5
3 7 9

sum(19)
8 10 1
6 5
2 4
3 7 9

sum(19)
9 7 3
4 2
5 6
1 10 8

sum(19)
9 7 3
4 6
5 2
1 10 8

sum(19)
9 7 3
5 2
4 6
1 10 8

sum(19)
9 7 3
5 6
4 2
1 10 8

sum(20)
9 7 4
3 1
6 5
2 8 10

sum(20)
9 7 4
3 5
6 1
2 8 10

sum(20)
9 7 4
6 1
3 5
2 8 10

sum(20)
9 7 4
6 5
3 1
2 8 10

sum(22)
9 7 6
1 3
2 5
10 4 8

sum(22)
9 7 6
1 5
2 3
10 4 8

sum(22)
9 7 6
2 1
3 5
8 4 10

sum(22)
9 7 6
2 3
1 5
10 4 8

sum(22)
9 7 6
2 5
1 3
10 4 8

sum(22)
9 7 6
2 5
3 1
8 4 10

sum(22)
9 7 6
3 1
2 5
8 4 10

sum(22)
9 7 6
3 5
2 1
8 4 10

sum(22)
10 4 8
1 2
5 3
6 7 9

sum(22)
10 4 8
1 3
2 5
9 7 6

sum(22)
10 4 8
1 3
5 2
6 7 9

sum(22)
10 4 8
1 5
2 3
9 7 6

sum(22)
10 4 8
2 3
1 5
9 7 6

sum(22)
10 4 8
2 5
1 3
9 7 6

sum(22)
10 4 8
5 2
1 3
6 7 9

sum(22)
10 4 8
5 3
1 2
6 7 9

sum(20)
10 6 4
2 1
5 7
3 9 8

sum(20)
10 6 4
2 7
5 1
3 9 8

sum(20)
10 6 4
5 1
2 7
3 9 8

sum(20)
10 6 4
5 7
2 1
3 9 8

sum(19)
10 8 1
2 5
3 7
4 9 6

sum(19)
10 8 1
2 5
4 6
3 9 7

sum(19)
10 8 1
2 6
4 5
3 9 7

sum(19)
10 8 1
2 7
3 5
4 9 6

sum(19)
10 8 1
3 5
2 7
4 9 6

sum(19)
10 8 1
3 7
2 5
4 9 6

sum(19)
10 8 1
4 5
2 6
3 9 7

sum(19)
10 8 1
4 6
2 5
3 9 7

sum(20)
10 8 2
1 3
5 6
4 7 9

sum(20)
10 8 2
1 6
5 3
4 7 9

sum(20)
10 8 2
5 3
1 6
4 7 9

sum(20)
10 8 2
5 6
1 3
4 7 9

Hope you all enjoyed the puzzle!