Repunits

Algebra Level 3

A repunit is a number that contains only the digit 1, and the repunit of n n digits is denoted by R n = 11111 … 1 ⏟ n times R_{n} = \underbrace{11111 \ldots 1}_{n \text{ times}} .

If f ( x ) = 90 x 2 + 20 x + 1 f(x) = 90x^2 + 20x + 1 , then find the sum of digits of the number f ( R 10 ) f(R_{10}) .

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The answer is 21.

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Utsav Banerjee by Utsav Banerjee , 30, USA

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1 solution

Utsav Banerjee
Apr 18, 2015

R n = 11111 . . . n t i m e s = 1 0 n − 1 + 1 0 n − 2 + . . . + 1 0 2 + 10 + 1 = 1 9 ( 1 0 n − 1 ) R_n=11111\;...\;n\;times=10^{n-1}+10^{n-2}+...+10^2+10+1 = \frac {1}{9} (10^n-1)

f ( R n ) = f ( 1 0 n − 1 9 ) = 90 ( 1 0 n − 1 9 ) 2 + 20 ( 1 0 n − 1 9 ) + 1 f(R_n)=f( \frac {10^n-1}{9})=90( \frac {10^n-1}{9})^2+20( \frac {10^n-1}{9})+1

⇒ f ( R n ) = 10 9 ( 1 0 n − 1 ) ( 1 0 n − 1 + 2 ) + 1 \Rightarrow f(R_n)=\frac {10}{9}(10^n-1)(10^n-1+2)+1

⇒ f ( R n ) = 10 9 ( 1 0 n − 1 ) ( 1 0 n + 1 ) + 1 \Rightarrow f(R_n)=\frac {10}{9}(10^n-1)(10^n+1)+1

⇒ f ( R n ) = 10 9 ( 1 0 2 n − 1 ) + 1 \Rightarrow f(R_n)=\frac {10}{9}(10^{2n}-1)+1

⇒ f ( R n ) = 1 9 ( 1 0 2 n + 1 − 10 + 9 ) \Rightarrow f(R_n)=\frac {1}{9}(10^{2n+1}-10+9)

⇒ f ( R n ) = 1 9 ( 1 0 2 n + 1 − 1 ) \Rightarrow f(R_n)=\frac {1}{9}(10^{2n+1}-1)

Therefore, f ( R n ) = R 2 n + 1 f(R_n) = R_{2n+1}

Hence, f ( R 10 ) = R 21 f(R_{10})=R_{21} , which has 21 ones, and the sum of its digits is 21.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Amazing solution. I did it the long way i.e. I did the big but simple multiplication. Good question. Keep posting

Aayush Patni - 6 years, 1 month ago

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