# Problems for the week

#### Problem 1.

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

#### Problem 2.

Find the minimum of if the equation has real roots.

#### Problem 3.

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

#### Problem 4. ### We will post the solutuions next week along with a new set of problem.

• Sounak Das

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

• No it’s not correct, your answer would have been correct when you can travel blocks north and 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 . My answer is a subset of these paths which are basically above the diagonal. Hope that clarifies your doubt.

• Sounak Das

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

Oo thanks.. yaa its clarify my doubt..

• Sounak Das

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

• yes, the 4 th problem answer is correct.

• ABHISHEK MISHRA

For the problem 2 answer is 17/4

• No, It’s • For P2 the answer is 4/5 I guess

• yes correct

• Ashish Shukla

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

• It’s . I will post the solution in 2 days.

• Ashish Shukla

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 !!!

• That was a brilliant answer. You are absolutely correct.