Divisible Subnumber Puzzle
Suppose that $ \overline{abcd} $ is a four-digit integer with no digits equal to zero such that $ \overline{ab} $, $ \overline{bc} $, and $ \overline{cd} $ are distinct integers that each divide into $ \overline{abcd} $. Find the smallest possible value of $ \overline{abcd} $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Since $ \overline{ab} | \overline{abcd} = 100 \cdot \overline{ab} + \overline{cd} $, then $ \overline{ab} $ also divides into $ \overline{abcd} - 100 \cdot \overline{ab} = \overline{cd} $. Similarly, since $ \overline{cd} | \overline{abcd} = 100 \cdot \overline{ab} + \overline{cd} $, then $ \overline{cd} $ must divide into $ \overline{abcd} - \overline{cd} = 100 \cdot \overline{ab} $. To minimize $ \overline{abcd} $, then we would like to try $ a = b = 1 $. It follows that $ \overline{cd} $ is divisible by $ 11 $, and also divides into $ 100 \cdot \overline{ab} = 1100 $. Thus, $ \overline{cd} = 11,22,44,55 $, but we can eliminate the first due to the distinctness condition. Trying each of the others, we see that $ 1122 = 2 \cdot 3 \cdot 11 \cdot 17 $ is not divisible by $ 12 $; $ 1144 = 2^3 \cdot 11 \cdot 13 $ is not divisible by $ 14 $; and $ \boxed{1155} = 3 \cdot 5 \cdot 7 \cdot 11 $ is indeed divisible by $ 15 $.
Freeflows