Grades 9-10 Video Solutions 2021-video 2021 9-10/23

video 2021 9-10/23

Video Transcription

Problem number 23 states, let n be the smallest positive integer, the sum of whose digits is 2,021. What is the sum of the digits of n plus 2,021?  So to start off the problem here, let's focus on the first part of the problem, which is finding out what the value of n is going to be.  So we know that n is going to be an integer with digits, we'll say a, b, c, on and on and on, all the way through x, y, z. We don't actually know how many digits n is going to have.  But keep in mind that this number is going to have a lot of digits. Why? Because we know that the sum of all of these digits must be equal to 2,021, and the largest  digit we can have is 9, which means there are going to be quite a few digits. So with these two restrictions, our goal is going to be to minimize n.  So let's take a look at some general rules that might help us achieve this minimum. First, let's use the least number of digits that we can.  For example, the sum of the digits 1 and 2, and 1,020 are both equal to 3, but 1,020 is a lot bigger because it has more digits.  Second, let's use the lowest value digits first because these digits are in the high place and will result overall in the lowest number.  So for example, 19 is going to be much less than 91. Keeping these rules in mind, let's try and find n.  By using our first rule that states that we should use as few digits as possible, let's try using the most of the highest digits we can in order to get to the target sum as fast  as possible. To do this, let's use a load of nines. We'll say that we use x nines. However, there's no guarantee that all these nines will add up to our target sum of 2,021.  So let's keep our second rule in mind and let's put in another digit. We'll call this digit a, and it will come at the front.  This will ensure that the sum of our digits is equal to 2,021. Now let's set our target sum equal to the sum of the digits.  In this case, since we know that a and 9 must be integers and a must be less than 10, we can divide 2,021 by 9 and get 224 with a remainder of 5.  That means that we have 224 nines and that a is equal to 5. Now for the second part of the question. What is the sum of the digits of our massive number in 2,021?  For this, let's do long addition. So let's just add normally how we normally would, starting off at the right. So 9 and 1 is equal to 10, carry the 1.  1 plus 9 plus 2 is 12, carry the 1. 1 plus 9 plus 0 is 10, carry the 1. 1 plus 9 plus 2 is 12, carry the 1. 1 plus 9 plus nothing is 10, carry the 1.  And here we have a lot of nines, so we're going to have 1 plus a lot of nines is going to be equal to 10.  So at the very end of this, we will have a 1 and at the bottom, we will have a continuation of zeros. 1 plus 9 is going to be 10, carry the 1.  1 plus 9 is going to be 10, carry the 1. And finally, we're going to have 1 plus 5, which is going to be 6. So now let's add up all of these digits.  So we're going to have 6 plus 0 plus 0 plus a lot of zeros plus 0 plus 2 plus 0 plus 2 plus 0. So in total, we just have a 6 and two 2s, which sum up to 10.  So the question asked us, what is the sum of the digits of n plus 2,021? The answer is 10, letter A.

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