Yet another wonderful numbers puzzle from Alok Mittal’s Mathematical Circles class. Not too difficult, and yet a lot of fun!
Take 10 cards of the same suite from Ace (A) through 10, i.e. essentially numbers 1-10. Using each card exactly once, construct a card frame (like the one shown below), where each side adds up to the same number.
Puzzle 116 Graphic
Happy arranging!
This was a relatively simple puzzle and the number of people who sent correct answers was overwhelming. I will try to capture as many as I can – Ishir Gupta (only one not to use algebra), Pratik Poddar, Suman Saraf, Kunal Goyal, Girish Tutakne, Karan Sharma, Abhinav Jain, Abhinav Kumar Gupta, Siddharth Goel, Chandu Karan, Manphul Budaniya, Ravi Kabra, Anupam Sarawagi and Garima Rai. Well done all – and thank you for your continued support and interest in this puzzle blog that keeps me going!
I like the clarity and detail in Abhinav Jain’s solution, though most people sent similar answers, but I am reproducing it (another answer later):
Let the no. be x
x>2000 (there is no such no. Between 1978-2000)
I am assuming x to be a 4 digit no.
1000a + 100b+ 10c + d = x
And
(10a + b) + (10c + d) = 10b + c
10a + d = 9(b-c)
Therefor 10a + d is a multiple of 9.
Where :
a isn’t equal to 0.
1<a<10
0<= b,c,d < 10
b > c
The possible values of 10a + d are :
1. 27 a=2 d=7
2. 36 a=3. d=6
3. 45. a=4. d=5
4. 54. a=5. d=4
5. 63. a=6. d=3
6. 72. a=7. d=2
7. 81. a=8. d=1
Since we have to find the immediate next no.
a=2 d=7
Therefor
b-c=3
We have to find the least value of b and c for which the equation is true
b=3
c=0
Therefor the no. Is 2307
And the next no. Will be 2417, 2527,2637………… The last 4- digit no. Will be
8901.
I also like the solution sent by Ishir, even though it is missing one step (of why there is a carry over), but I really like the thinking and it is done without algebra:
This is a puzzle I got recently from my daughter, Anisha, who in turn got this puzzle from Alok Mittal’s Mathematical Circles Class.
There is an island on which a rabbit arrives. Rabbit reproduces, as part of which it gives birth to other rabbits. No rabbit dies on the island and not all rabbits have to reproduce. The way they reproduce is as follows:
No rabbit has died and not all rabbits have to reproduce
Grey – 40; spotted 30 and White is 40
What color is the first rabbit?
Happy figuring out!
I got an overwhelming response to this puzzle – so many that putting down the names of correct answers is not feasible. This was also the best viewership of the blog for any week since I started this blog – thanks a ton!
The answer to the puzzle is as follows:
7 A Q 2 8 3 J 4 9 5 K 6 10
Here is the link to the answer:
https://www.educreations.com/lesson/view/solution-to-puzzle-113/33322903/?s=rekB2E&ref=app
The primary trick was to think of the cards as being in a circle and then begin to put the cards in the space as they show up.
Hope you all enjoyed the puzzle!
This is a puzzle for the number buffs that I found on http://www.mindcipher.com
The year 1978 is such that the sum of the first two digits and the latter two digits is equal to the middle two digits, i.e. 19 + 78 = 97. What is the next year (after 1978) for which this is true? (please do not write computer programs to solve)
Please send your answers either directly on the blog site as comments, or to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.
Happy year hunting!
As I mentioned in the question, this is an age old problem, but I was still intrigued by the twist provided to this puzzle by a colleague’s daughter. Many people gave a correct answer to the first two parts of the puzzle, only Pratik Poddar came back with the right answer for the 3rd part.
Here are the answers (and I will add explanations later):
Part 1: Weights need to be 1, 2, 4, 8, ….essentially all the powers of 2
Part 2: Weights need to be 1, 3, 9, 27, 81….essentially powers of 3
Part 3: Weights need to be twice that of powers of 3, i.e. 2, 6, 18, 54, ….
The basic principle to be adopted is that you choose weights in a manner, so that no weight can be represented in two different ways. For example, if you had two weights of 5 kg each, then you can make 10 kg with it (in Part 1). However, there is a bit of “wastage” here as well, as 5 kg itself can be formed by two different weights.
In Part 1, if you begin with a weight of 1 kg, you can only weigh 1 kg with it. For 2 kg, you need to add a weight of 2 kg. With the two of them, you can weigh 3 kg also, but not 4 kg, so you add a weight of 4 kg. Now with these three weights, you can go up to 7 kg, but not 8 kg…you can continue this way and realise that if you have weights that are a power of 2, you can create any weight with these combinations, by effectively using the binary notation.
In a balance, you now have a special property, that you can put weights on both sides and therefore not only add weights, but also subtract weights. So lets begin again:
– you start with a weight of 1 kg
– Add a weight of 2 kg. Now you can measure 2 kg with it. However, you can now create 1 kg in two different ways – 1 kg by itself and also by putting 2 kg on one side and 1 kg on the other side (difference). That is “wastage”
– So instead of 2 kg, we try 3 kg. With 3 kg, you can measure 2 kg through a difference of 3 kg and 1 kg. You can also measure 3 kg, and also 4 kg. But you cannot measure 5 kg.
– Through trial and error, you will realise that there is wastage by trying 5 kg, 6 kg, 7 kg and 8 kg. Once you try 9 kg, you can measure 5-8 kg by subtracting 1-4 kg from 9 kg.
– A better way to think about this is – since you can measure all the weights up to 4 kg, the next weight you need to chose is 4 x 2 + 1 (9 kg), because all the intermediate weights can be measured by subtracting from 9. With these weights, you can measure all the way up to 13, and hence the next weight needs to be 2 x 13 + 1 = 27 kg…and so on.
Part 3: The trick here is that once you know that whatever you are trying to measure is heavier than lets say 15 kg and lighter than 17 kg, and you know that it can only be a whole number, then you know that it is 16 kg, without actually having a combination of weights that gives you 16. This is what Arushi Gupta pointed out.
When you think about this further, you can conclude that if you can measure all alternate numbers, then the ones inbetween can be inferred. The easiest way to do this is from Part 2, you know that you can measure all #s, so by multiplying the weights by 2, you can measure 2, 4, 6, 8, 10, ….and infer the odd ones!
Hope you enjoyed the puzzle!
This is a puzzle I have known for a very long time, and do not remember who asked me, I probably did it when I must have been around 15 years old. I loved this puzzle of cards. Unlike other puzzles, this requires me to show a video, which I have uploaded on YouTube.
Take out all the cards of one single suite in a pack of 52 cards e.g. take out all the cards of spade (13 cards). Your task is to arrange them in a order (face down) so that if you move the top card at the bottom of the pack and then put down face up the card that shows up next, and continue this in a sequence, then you will get all the cards from A, 2, 3, …., K.
To demonstrate what I am talking about, please look at the following video:
Please send your answers either directly on the blog site as comments, or to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.
Happy arranging!
This was a difficult puzzle. I had trouble understanding the answer even when it was given to me, and had to go to Alok Mittal for rescue. Siddharth Goel (Founder, Adstrak) gave some useful links for this puzzle and Sumar Saraf was the only one to give the right answer, though with a bit of brute force and not the actual proof.
I am reproducing the answer sent to me by Alok Mittal.
Mark every person by two numbers – the max number of consecutive ascending chain that ends at him or her, and max descending chain that again ends at him or her (beginning from the left). No two people can have same (asc,des) pair – the reason being that both have different heights. If the one standing on the right has a smaller height then the des # for the right one will at least be one higher than the one standing on the left and vice versa for the asc number in case the right one is taller. We also know that the minimum number for sac or des needs to be at least 1. Hence with 10 people, someone must have a “4” in at least one of ascending or descending tag.
In general, for numbers between n^2+1 to (n+1)^2, the minimum will be n+1.
A more general treatment of the problem is available on Wikipedia at the following link:
https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Szekeres_theorem
Hope you enjoyed the puzzle!
This is a twist to a puzzle which has been around for a very long time. I have myself looked at this puzzle since I was a kid, but a twist (or a potential flaw) to the puzzle was recently pointed out by Arushi (a 9th grader), a colleague’s daughter based in California. I was very impressed with Arushi’s thinking and am taking the liberty of sharing this with everyone. I am articulating the puzzle at three different levels (of varying difficulty):
Please send your answers either directly on the blog site as comments, or to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.
Happy balancing acts!
This was a relatively easy puzzle and many people sent a correct answer. That included Siddhartha Goel, Suman Saraf, Pratik Poddar, Traveling Salesman (which I believe is a nick name), Narsimha Pai and Salil Panikkaveettil.
The puzzle used a popular and intuitive principle called the Pigeon Hole principle. At a simplistic level, it states that if n items are put into m containers, with n > m, then at least one container must contain more than one item. This surprisingly simple principle is used very widely and finds many applications.
I am taking the liberty of reproducing Siddhartha Goel’s answer for this puzzle:
1) n – eaters, n – entrees
2) no duplicate entrees etc: each user has a 1:1 match with 1 entree
3) A permutation can be rotated ‘n’ times and a total of n matches (eater – entree) will happen in ‘n’ rotations
4) If in 1 permutation we have 0 matches => in remaining (n-1) rotations of that permutation there needs to be a total of n matches => at least 1 rotation of the permutation will have at least 2 matches… (0, 1, 1, 1, … 2)…
For people, who are interested in a more detailed treatment of Pigeon Hole principle, please refer to https://en.wikipedia.org/wiki/Pigeonhole_principle
Hope you enjoyed the puzzle!
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