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

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smecloudbird18 smecloudbird18 · Answer 1 · 2022-11-30T08:15:56+01:00

Sum of all positive integers having twice as many digits when written in base $2$ as they have when written is base $3$ is 10.

This is because the only two integers which fits the constraint are 2 and 8,

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

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

The result of the expression $a+b$ is 11

This is because the LCM of [1!+2!, 2!+3!, 3!+4!, 4!+5!, 5!+6!, 6!+7!, 7!+8!, 8!+9!] =3,628,800

Which can be written as : 1. 10! = 1 + 10 = 11

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