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Number Theory Level 2

y,z are digits such that p= 0.yzyzyzyz….. q=0.yz r=0.zyzyzy……. If 275p=1+275q, possible value of r-p is/are equal to

note:- first two digit is enough


The answer is 0.27.

Vishal S by Vishal S , 21, India

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1 solution

Vishal S
Jan 6, 2015

Given that

p=0.yzyzyz... =0. y z ‾ \overline{yz}

q=0.yz

r=0.zyzyzy... =0. z y ‾ \overline{zy}

When we simplify the given equation, we get

275p=1+275q

⇒ \Rightarrow 275(p-q)=1

⇒ \Rightarrow 275(0.yzyzyz..-0.yz)=1

⇒ \Rightarrow 275(0.00yzyz...)=1

⇒ \Rightarrow 0.00yzyz...= 1 275 \frac {1}{275}

By multiplying 100 on both the sides, we get

⇒ \Rightarrow 0.00yzyz.. x 100= 1 275 \frac {1}{275} x 100

⇒ \Rightarrow 0.yzyz..= 100 275 \frac {100}{275}

⇒ \Rightarrow 0.yzyz..=0.3636...

⇒ \Rightarrow y=3, z=6

*By substituting y=3 & z=6 in p & r we get *

r-p=0.636363.. - 0.363636......

⇒ \Rightarrow r-p=0.2727.. =0. 27 ‾ \overline{27}

Therefore the value of r-p=0. 27 ‾ \overline{27}

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