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Tuesday, July 7, 2015

An Amazing Level 3 Logic Problem from Brilliant That Could Be Solved by 29% of Those Who Attempted It (On Brilliant)

I am going to be out of station from 11th July to 15th July.So,I'll miss posting on these weekends but can I let it happen?No!I have been so excited to share the problems I have been solving this week.I will be sharing two more within tomorrow before I leave the town.Loved this logic problem.

I have a pack of pens to distribute. If I keep 4, 5 or 6 in a pack, I am left with 3 pens. If I keep 7 in a pack, I am left with none. What is the minimum number of pens I have to pack and distribute?Read on to find out the detailed solution.

Moreover,I saw a Massachusetts Institute of Technology Graduate in the list of 11 solvers who have been able to do the sum on Brilliant. 

Let the total number of pens be n.
According to the question, n = 4x + 3.Again,n= 5y + 3 and n = 6z+3,where x,y and z are unknown variables.Now,n = 7t as well,where t is a variable.
Look at the ones digit now.If you add 3 to the ones digit of the possible multiples of 4 (since the remainder is 3),you get 7,1,5,9 or 3.
Similarly,if you add 3 to the multiples of 5,you may get only 3 or 8.
If you do the same to the multiples of 6,you may get 9,5,1,7 and 3.
So,the only common digit is 3 and the minimum number with 1 digit 3 in its one's place which is divisible by 7 is 63.So,the minimum number of pens has to be 63.
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  1. Finally I found a girl who is damn good in maths. Otherwise you know how girls are with maths ;)

    1. Thanks for the compliment. :)
      Actually most of them just do not like to pay attention but we are rising.Look at the IAS toppers (Top 3) this year.And even the MIT student I am talking about is a female.I am proud of all women who are meritorious,strong,courageous and honest. :)


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