1867 is multiplied by a positive integer k. the last four digits of the product are 2017. What is the number of digits in the smallest possible value of k?

Question

This is a Math Olympiad question for high grades and i am only 6th grade.

Answer 1

This problem gives you an excellent opportunity to start with a situation and to ask, "Now, how did we get here?"  Many, many jobs require that skill.

 

When 1867 is multiplied by kkkk to get ddddd2017, we know that 1867 divided into ddddd2017 gives kkkk.  We also know that the ones place of the product must be the ones place of 7*kkkk:

       1867

    *  kkkk

      ----------

     ?????X      [X is ones digit of 7*k, possibly with a carry]

 

Now, the multiples of 7 are:

    0   7   14   21   28   35   42   49   56   63   etc. (note: digits in ones place repeat)

 

The only way to get 7 in the ones place of the product is to have 1 in the ones place of kkkk (that is, kkk1).

 

Since there was no carry, how do we get 1 (from 2017) in the tens place of the product?  Remember that we already have a 6 (from 1867) there, so 11-6=5 and we must find a multiple of 7 that ends with 5.  In the list above, we find that 7*5=35.  The value kkkk must be kk51.

 

See if you can determine how to use a carry of 3 with the numbers above to get a 0 in the hundreds place of the product:

          1867

        * kk51

      -----------

          1867

         9335

     ####                [we need this to be ###8, so kkkk must be kkk451]

      -----------

      ##2017

 

Now, we have:

          1867
        * k451
      -----------
           1867
         9335

        7468

   #####                           [we need this to be ####0, so kkkk is 0451]

      -----------

    ###2017

 

Important note: the digits of kkkkkk beyond 0451 are not important.  Besides, the problem asks for the lowest value of k that produces a product with 2017 as the last four digits.

 

So, k=451 because 1867*451 = 842017