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

#2**+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