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What is the sum of all positive integers that have twice as many digits when written in base $2$ as they have when written in base $3$? Express your answer in base $10$.

 

The least common multiple of $1!+2!$, $2!+3!$, $3!+4!$, $4!+5!$, $5!+6!$, $6!+7!$, $7!+8!$, and $8!+9!$ can be expressed in the form $a\cdot b!$, where $a$ and $b$ are integers and $b$ is as large as possible. What is $a+b$?

 May 25, 2018
 #1
avatar+102765 
+2

I think...

 

\(f^{-1}\left(\frac{13+2}{5}\right) = f^{-1}(3) =1\)

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 May 25, 2018
 #2
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+2

What is the sum of all positive integers that have twice as many digits when written in base $2$ as they have when written in base $3$? Express your answer in base $10$.

 

2 in base 10 =10 in base 2 = 2 in base 3

8 in base 10 =1000 in base 2 = 22 in base 3.

2 + 8 = 10 the sum of positive integers.

 

The least common multiple of $1!+2!$, $2!+3!$, $3!+4!$, $4!+5!$, $5!+6!$, $6!+7!$, $7!+8!$, and $8!+9!$ can be expressed in the form $a\cdot b!$, where $a$ and $b$ are integers and $b$ is as large as possible. What is $a+b$?

 

LCM[1!+2!, 2!+3!, 3!+4!, 4!+5!, 5!+6!, 6!+7!, 7!+8!, 8!+9!] =3,628,800. This can be written as:

1. 10! = 1 + 10 = 11, which is a + b

 May 25, 2018
edited by Guest  May 25, 2018

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