 # Understand the problem

The product $(8)(888\dots8)$, where the second factor has $k$ digits, is an integer whose digits have a sum of $1000$. What is $k$?

##### Source of the problem
American Mathematics Competition
Number Theory

7/10

##### Suggested Book

Elementary Number Theory by David M, Burton

So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further !

Think you could do with some help ? Okay let’s get ourselves a headstart. As you might have guessed, sometimes the most fruitful thing to be done is to observe what’s going on in these kind of problems. The intuition is clear, there are too many instances of ‘8‘-s for someone to account for them manually.  So, yes, that’s something we can expect. Try working out with a few simple cases like, k=1, k=2, and so on…

Okay, so let’s see what’s happening for a few small values of k…
k=1
8 * 8 = 64 k=2
8 * 88 = 704 k=3
8 * 888 = 7104 k=4
8 * 8888 = 71104 …
So, Wait ! Look closely…Do you see something ?

A really nice pattern is evolving. If you can see it, and yet find it a bit difficult to articulate it mathematically, don’t worry. See for k=4, the product gives us the result 71104. That means, we have the starting digit to be 7. And the ending suffix is 04. What varies are the ones. See, for k=4, we have ( k-2 = 2 ) ones. That’s it ! The generalization should be fairly simple for us to do now… For every k >=2, the product result consists of a 7 to start with, exactly k-2 1’s follow, and we conclude with single occurrences of 0 and 4 each. Now, think…can you take this till the end ?

Once we’ve seen the pattern, it’s easy to get this done.  What we have noted, we need that to sum up to 1000.  In simple mathematical terms, 7 + (k-2).1 + 0 + 4 = 1000
11 + k = 1002
k = 991

So, we need ‘8’ to come 991 times in the multiplicand, so that the digits sum up to 1000. So, that seals the deal !

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## Coin Toss Problem | AMC 10A, 2017| Problem No 18

Try this beautiful Problem on Probability from AMC 10A, 2017. Problem-18, You may use sequential hints to solve the problem.

## GCF & Rectangle | AMC 10A, 2016| Problem No 19

Try this beautiful Problem on Geometry on Rectangle from AMC 10A, 2010. Problem-19. You may use sequential hints to solve the problem.

## Fly trapped inside cubical box | AMC 10A, 2010| Problem No 20

Try this beautiful Problem on Geometry on cube from AMC 10A, 2010. Problem-20. You may use sequential hints to solve the problem.

## Measure of angle | AMC 10A, 2019| Problem No 13

Try this beautiful Problem on Geometry from AMC 10A, 2019.Problem-13. You may use sequential hints to solve the problem.

## Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Geometry from PRMO -2018.You may use sequential hints to solve the problem.

## Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

## Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.

## Set of Fractions | AMC 10A, 2015| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2015. Problem-15. You may use sequential hints to solve the problem.

## Indian Olympiad Qualifier in Mathematics – IOQM

Due to COVID 19 Pandemic, the Maths Olympiad stages in India has changed. Here is the announcement published by HBCSE: Important Announcement [Updated:14-Sept-2020]The national Olympiad programme in mathematics culminating in the International Mathematical Olympiad...

## Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2015. Problem-24. You may use sequential hints to solve the problem.