ABBABB

Number Theory Level 2

The product of six consecutive prime numbers is equal to $\overline{ABBABB}$ where $A$ and $B$ are two distinct positive digits. What is the value of $A+B$ ?

Note: $\overline{ABBABB}$ is in base $10$ .

12 5 7 8 11 9 6 10

1 solution

Áron Bán-Szabó
Jul 13, 2017

$\begin{aligned} \overline{ABBABB} & = 100000A+10000B+1000B+100A+10B+B \\ & = 100100A+11011B \\ & =1001*(100A+11B) \\ & = 7*11*13*(100A+11B) \end{aligned}$

Since this number is equal to the product of six consecutive prime numbers and $2*3*5=30<100A+11B<17*19*23=7429$ , both of the $5$ and $17$ is a divisor of $(100A+11B)$ . The $5*17=85$ has only one multiple which can be expressed in a $\overline{ABB}$ formula, the $255$ , so $A=2$ and $B=5$ , and $255255=3*5*7*11*13*17$ .

Therefore the answer is $2+5=\boxed{7}$ .

It would be easier to just say "Since 7, 11, 13 are 3 of the prime factors, we just have to check the values of $2 \times 3 \times 5, 3 \times 5 \times 17, 5 \times 17 \times 19, 17 \times 19 \times 23, \ldots$ .

Calvin Lin Staff - 3 years, 11 months ago

ABBABB

1 solution

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