WEEK 3 CHALLENGING PROBLEMS

Fractions > WEEK 3 CHALLENGING PROBLEMS

Problems for the week

Deadline 03.05.2020

Problem 1.

Can the product of 3 consecutive integers be a power of an integer? What about 4 consecutive integers?

Problem 2.

Find the minimum of a^2+b^2 if the equation x^4+ax^3+bx^2+ax+1=0 has real roots.

Problem 3.

You walk n blocks north and n blocks east of your home to reach your school. Every day you have 2n blocks to walk. How many such shortest routes are possible if you never cross the diagonal line from home to school?

Problem 4.

\Large{\int_{0}^{\frac{\pi}{4}}x\prod_{k=1}^{\infty}\cos\bigg(\frac{x}{2^k} \bigg) dx=?}

To check week 1 challenging problems and discussion: Click here

We will post the solutuions next week along with a new set of problem.

14 Comments

  • Sounak Das
    April 28, 2020 at 11:23 am Reply

    Problem-3.. Since we are required the shortest possible path..we can’t go backward and downward direction. We can only go forward& upward direction. We can choose 2nCn for north blocks..rest blocks must be east block automatically. So the ans is 2nCn.

    I guess its the correct explanation& ans

    • FractionsHub
      April 28, 2020 at 2:32 pm Reply

      No it’s not correct, your answer would have been correct when you can travel n blocks north and n blocks east with no restriction of crossing the diagonal. Suppose you are traveling from (0,0) to (n,n) so you can first take the n east blocks and then take the n north blocks, but this path is below the diagonal, and so not considered in my case. All shortest paths from (0,0) to (n,n) with no restriction is \binom{2n}{n}. My answer is a subset of these paths which are basically above the diagonal. Hope that clarifies your doubt.

  • Sounak Das
    April 28, 2020 at 6:44 pm Reply

    Problem-1. .
    (n-1)n(n+1)=x^m
    (n^2-1)n=x^m..Now, gcd((n^2-1),n) =1 .So take n^2-1=a^k, n=b^k ..
    Now,a^k – b^2k=1 . there’s no such two power whose difference is 1.
    Again suppose n(n+1)(n+2)(n+3)=x^m..
    (n^2+3n)(n^2+3n+2)=x^m
    gcd((n^2+3n)/2, (n^2+3n+2)/2)=1
    So, we can again write
    (n^2+3n)/2=a^k & (n^2+3n+2)/2=b^k ( we can write it because x^m is divisible by 4)
    Again two power difference of two positive integer have difference 1.

  • Sounak Das
    April 28, 2020 at 6:46 pm Reply

    Oo thanks.. yaa its clarify my doubt..

  • Sounak Das
    April 28, 2020 at 6:48 pm Reply

    For problem 4 is the answer is 1-1/√2?

  • ABHISHEK MISHRA
    May 5, 2020 at 2:30 pm Reply

    For the problem 2 answer is 17/4

  • Uditanshu Sadual
    May 7, 2020 at 12:29 pm Reply

    For P2 the answer is 4/5 I guess

  • Ashish Shukla
    May 13, 2020 at 2:20 pm Reply

    I want to know the answer of Question no. 2 of week- 3
    Is it – 12 ??
    I solved it by converting bi-quadratic into quadratic by dividing whole equation by x^2 and then applying the conditions of real root for a quadratic equation i.e D≥0 and I got
    a^2-4b+8≥0
    or, a^2+b^2 ≥ b^2+4b-8(on adding b^2 to both sides)
    or, min. of (a^2+b^2) = min of ( b^2+4b-8)
    = min of {(b+2)^2 – 12 }
    At b = -2 , we get the min
    Which is -12
    Please reply

  • Ashish Shukla
    May 13, 2020 at 7:01 pm Reply

    Yes ,answer is 4/5
    I realized my mistake
    Actually , the equation reduces upto
    t^2+at+b-2 = 0 where t=x+1/x
    Now ,
    As x≥0 , x+1/x ≥ 2 (using AM-GM inequality)
    ஃ t ≥ 2
    On solving quadratic ‘t’ comes out to be
    t = -a ± {a^2- 4(b-2)}^1/2 / 2
    or , -a ± {a^2- 4(b-2)}^1/2 ≥ 4
    or , {a^2- 4(b-2)}^1/2 ≥ a+4
    or , a^2 – 4b + 8 ≥ a^2 + 8a +16(on squaring)
    or , – b ≥ 2a + 2
    or, b^2 ≥ 4a^2 + 8a + 4 (on squaring )
    or , a^2 + b^2 ≥ 5a^2 + 8a + 4 (adding a^2 to both sides )

    or , a^2 + b^2 ≥ {( 5a^2 + 8a + 4)*5}/5 ( on multiplying numerator and denominator by 5)

    or , a^2 + b^2 ≥ {(25a^2 + 40a + 16 + 4)}/5

    or , a^2 + b^2 ≥ {(5a+4)^2 + 4}/5
    or , a^2 + b^2≥4/5 ( at, a = -4/5)
    That’s all !!!

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