An Infinite Number Of Xs

Level 2

If: 7 = X 11 + X 121 + X 1331 + X 14641 + t o 7=\frac{X}{11}+\frac{X}{121}+\frac{X}{1331}+\frac{X}{14641}+\dotsm\;to\;\infty Find the value of X X


The answer is 70.

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5 solutions

Let 7 = S 7=S .Then: S = X 11 + X 121 + X 1331 + X 14641 + S=\frac{X}{11}+\frac{X}{121}+\frac{X}{1331}+\frac{X}{14641}+\dotsm S 11 = X 121 + X 1331 + X 14641 + \frac{S}{11}=\frac{X}{121}+\frac{X}{1331}+\frac{X}{14641}+\dotsm S S 11 = ( X 11 + X 121 + ) ( X 121 + X 1331 + ) S-\frac{S}{11}=\left(\frac{X}{11}+\frac{X}{121}+\dotsm\right)-\left(\frac{X}{121}+\frac{X}{1331}+\dotsm\right) 10 S 11 = X 11 \frac{10S}{11}=\frac{X}{11} 10 S = X 10S=X Plugging in S = 7 S=7 ,We get: 10 ( 7 ) = X = 70 10(7)=X=\boxed{70}

Why in the final, S = 7 S=7 ?

. . - 4 months ago

A nice solution, but be wary of manipulating infinite sums like that

see http://tinyurl.com/offobv8

Luka Srot - 6 years, 6 months ago
. .
Feb 9, 2021

7 = X 1 1 1 + X 1 1 2 + X 1 1 3 + X 1 1 4 + n = 1 , 7 = X 1 1 n + X 1 1 n + 1 + X 1 1 n + 2 + 7=\frac{X}{11^1}+\frac{X}{11^2}+\frac{X}{11^3}+\frac{X}{11^4}+\cdots \rightarrow n =1, 7 = \frac{X}{11^n}+\frac{X}{11^{n+1}}+\frac{X}{11^{n+2}}+\cdots . Then lets get a approximation of 7 = X 11 7 = \frac{X}{11} . Maybe it is 77 = X 77=X . Next, 7 = X 1 1 2 847 = X 7=\frac{X}{11^2} \rightarrow 847=X . And so on.... Then, let the final number A A . Next, A = X 1 1 b , b 1 A=\frac{X}{11^{b}}, b \geq 1 . Then 7 × 1 1 b 1 = ( b 1 ) X + ( b 2 ) X + ( b 3 ) X + 7 \times 11^{b-1} = (b-1)X+(b-2)X+(b-3)X+\cdots . Although we do not know the value of b b , but lets keep calculating. In 7 = X 11 + X 121 + X 1331 + X 14641 7 = \frac{X}{11}+\frac{X}{121}+\frac{X}{1331}+\frac{X}{14641} , we get 102487 = 1331 X + 121 X + 11 X + X 102487 = 1331X + 121X + 11X + X . Then 1464 X = 102487 1464X = 102487 . Lets finish the calculating with X = 102487 1464 X = \frac{102487}{1464} . Lets round it into the integer form. Its approximation is 70 70 , so the answer is 70 \boxed{70} .

Jack Rawlin
Feb 11, 2016

n = 0 a r n = a 1 r \sum_{n = 0}^{\infty}{ar^n} = \frac{a}{1 - r}


7 = X 11 + X 121 = X 1331 + X 14641 + 7 = \frac{X}{11} + \frac{X}{121} = \frac{X}{1331} + \frac{X}{14641} + \cdots

a = X 11 , r = 1 11 a = \frac{X}{11},~ r = \frac{1}{11}


n = 0 A x ( B x ) n = A x B \sum_{n = 0}^{\infty}{\frac{A}{x} \cdot \left(\frac{B}{x}\right)^n} = \frac{A}{x - B}


A = X , B = 1 , x = 11 A = X,~ B = 1,~ x = 11

n = 0 X 11 ( 1 11 ) n = X 11 1 = X 10 \sum_{n = 0}^{\infty}{\frac{X}{11} \cdot \left(\frac{1}{11}\right)^n} = \frac{X}{11 - 1} = \frac{X}{10}

7 = X 10 , X = 70 7 = \frac{X}{10},~ X = 70

X = 70 \large \boxed{X = 70}

Aareyan Manzoor
Jan 2, 2015

7 = X ( 1 11 + 1 121 + 1 1331 + 1 14641 + ) 7=X(\frac{1}{11}+\frac{1}{121}+\frac{1}{1331}+\frac{1}{14641}+\dotsm) s = ( 1 11 + 1 121 + 1 1331 + 1 14641 + ) s=(\frac{1}{11}+\frac{1}{121}+\frac{1}{1331}+\frac{1}{14641}+\dotsm) 11 s = 1 + 1 11 + 1 121 + 1 1331 + 1 14641 + 11s=1+\frac{1}{11}+\frac{1}{121}+\frac{1}{1331}+\frac{1}{14641}+\dotsm 11 s = 1 + s s = 1 10 11s=1+s \longrightarrow s=\dfrac{1}{10} 7 = X s 7 = X 10 X = 70 7=Xs \longrightarrow 7=\dfrac{X}{10} \longrightarrow X=\boxed{70} note that n = 1 1 S n = 1 S 1 \sum_{n=1}^\infty \dfrac{1}{S^n}=\dfrac{1}{S-1}

Ninad Akolekar
Nov 16, 2014

Quite easy. Just use the formula for infinite sum of a GP.

So what is it? I let you explain that.

. . - 4 months ago

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