+0  
 
0
165
2
avatar+162 

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 $a+b$?

Creeperhissboom  May 25, 2018
 #1
avatar+92888 
+2

I think...

 

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

Melody  May 25, 2018
 #2
avatar
+1

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

Guest May 25, 2018
edited by Guest  May 25, 2018

22 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.