If x=cos1cos2cos3.....cos89 and y=cos2cos6cos10....cos86, then what is the integer nearest to \(\frac{2}{7}log_2{\frac{y}{x}}\)?

\(\frac{y}{x}\)=\(\frac{cos2cos6cos10.....cos86}{cos1cos2cos3....cos89}\)

=\(2^{44}\times\sqrt{2}\frac{cos2cos6cos10...cos86}{sin2sin4...sin88}\)

[ since cos\(\theta\)=sin(90-\(\theta\)) from cos 46 upto cos 89 and 2sin\(\theta\)cos\(\theta\)=sin2\(\theta\)]

=\(\frac{2^{\frac{89}{2}}sin4sin8sin12...sin88}{sin2sin4sin6...sin88}\)

[ since sin\(\theta\)=cos(90-\(\theta\))]

=\(\frac{2^{\frac{89}{2}}}{cos4cos8cos12..cos88}\)

[ since cos\(\theta\)=sin(90-\(\theta\))]

=\(\frac{2^\frac{89}{2}}{\frac{1}{2}^{22}}\)

[since \(cos4cos8cos12...cos88\)

\(=(cos4cos56cos64)(cos8cos52cos68)(cos12cos48cos72)(cos16cos44cos76)(cos20cos40cos80)(cos24cos36cos84)(cos28cos32cos88)cos60\)

\(=(1/2)^{15}(cos12cos24cos36cos48cos60cos72cos84)\)

\(=(1/2)^{16}(cos12cos48cos72)(cos24cos36cos84)\)

\(=(1/2)^{20}(cos36cos72)\)

\(=(1/2)^{20}(cos36sin18)\)

\(=(1/2)^{22}(4sin18cos18cos36/cos18)\)

\(=(1/2)^{22}(sin72/cos18)\)

\(=(1/2)^{22}\)]

=\(2^\frac{133}{2}\)

\(\frac{2}{7}log_2{\frac{y}{x}}\)=\(\frac{2}{7} \times \frac{133}{2}\)=19.

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Consider all 6-digit numbers of the form abccba where b is odd. Determine the number of all such 6-digit numbers that are divisible by 7.

abccba (b is odd)

=a(\(10^5\)+1)+b(\(10^4\)+10)+c(\(10^3\)+\(10^2\))

=a(1001-1)100+a+10b(1001)+(100)(11)c

=(7.11.13.100)a-99a+10b(7.11.13)+(98+2)(11)c

=7p+(c-a) where p is an integer

Now if c-a is a multiple of 7

c-a=7,0,-7

hence number of ordered pairs of (a,c) is 14

since b is odd

number of such number=\(14 \times 5\)=70.

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Let a and b natural numbers such that 2a-b, a-2b and a+b are all distinct squares. What is the smallest possible value of b?

2a-b=\(k_1^2\) is equation 1

a-2b=\(k_2^2\) is equation 2

a+b=\(k_3^2\) is equation 3

adding 2 and 3 we get

2a-b=\(k_2^2+k_3^2\)

or, \(k_2^2+k_3^2\)=\(k_1^2\) \((k_2<k_3)\)

For least 'b' difference of \(k_3^2\) and \(k_2^2\) is also least and must be multiple of 3

or, \(k_2^2\)=a-2b=\(9^2\) and \(k_3^2\)=a+b=\(12^2\)

or, \(k_3^2-k_2^2\)=3b=144-81=63

or, b=21

or, least b is 21.

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