We can decompose the eeeeee factor first: eeeeee = e *1111= e * 3 * 7 *1*.
If a is 3e, abcde is 7 *11*13 * 37 = 37037, where a=3, which is substituted into the original formula, 37037 * 3 = 165438.
Then when A is E, then abcde is 3 * 7 *1*13 * 37 =11,where A =1/kloc-0.
When a=3, then ABCDE = 7 *1*13 * 37 * e =15873 * e, and a=3, then e can only be equal to 2, so substitute it into the original formula, ABCDE =/kloc.
When a=7e, ABCDE = 3 *11*13 * 37 =15873, where a= 1 and a=7e. Obviously, the original formula is not valid.
When a=7, then abcde= 15873*e, because a=7, so E can only be equal to 5. Substitute the original formulas abce= 15873*5=79365, 79365 * 7 = 55555. The original formula holds. So the final answer is 79365*7=555555.
Having said so much, I don't know if you understand. Of course, this is an example of all the possibilities. In fact, in the exam, you just need to learn the method and substitute the simplest one. We divide this learning into learning *1111= learning * 3 * 7 *1*13 * 37, and there is only 3 here. That's easy!