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# Understand the problem

The product \$$(8)(88888……8)\$$, where the second factor has k digits, is an integer whose digits have a sum of \$$1000\$$. What is k? $\\textbf{(A)}\\ 901\\qquad\\textbf{(B)}\\ 911\\qquad\\textbf{(C)}\\ 919\\qquad\\textbf{(D)}\\ 991\\qquad\\textbf{(E)}\\ 999$

##### Source of the problem
American Mathematical Contest 10A Year 2014

Number Theory

7/10

##### Suggested Book

Problem Solving Strategies Excursion In Mathematics

Do you really need a hint? Try it first!

After having a long look into this problem you can first make attempt by listing the first few numbers of the given form.Give it a try!!!!!

So we can do it like this 8*(8)=64 8*(88)=704 8*(888)=7104 8*(8888)=71104 8*(88888)=711104 Now try to observe the pattern in the above table because here lies the main insight of this problem . Come on cook it up!!!!!!

So form the table you can observe the terms are following a pattern that’s is The first number is 7 Then k-2 number of 1 Then the last two digits are 04

Now try to make the sum to 1000

So now you are in the final part so you can easily find 7+04+(k-2)=1000

implies 11+(k-2)=1000 . Solving this equation we get the value of K is 991 which is the required answer.

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