Question

6. The list 9, 18, 27, 36, … consists of all positive integers whose digit sum is 9 in increasing order. What is the 2014th number in this list? Solution:1000125 /Help me please/

Answer 1

To find the 2014th number in the list, we first need to find the pattern in the sequence. Notice that the first few numbers are:

9

18

27

36

45

54

63

72

81

90

108

117

126

135

144

153

162

171

180

...

We can see that every multiple of 9 is included in the sequence (i.e. 9, 18, 27, 36, etc.), as well as other numbers whose digits sum to 9 (such as 45, 54, 63, etc.). To find the 2014th number in the sequence, we can use the formula:

nth term = 9n + d

where n is the position of the term in the sequence and d is the difference between the sum of the digits in the nth term and 9. For example, the 12th term in the sequence is 108, which has a digit sum of 9 and therefore d=0. Plugging in n=12, we get:

nth term = 9(12) + 0

nth term = 108

To find the 2014th term, we need to solve the equation:

9n + d = 2014

Since we know that every multiple of 9 is included in the sequence, we can start by finding the largest multiple of 9 that is less than or equal to 2014:

9 x 223 = 2007

So we know that the 2014th term is somewhere between the 223rd multiple of 9 and the 224th multiple of 9. To narrow it down further, we can look at the difference between the current sum (d) and 9. Since the difference between any multiple of 9 and 9 is 0, we only need to look at the other terms that have a digit sum of 9. We can write out the terms as follows:

9 x 223 + 9

9 x 223 + 18

9 x 223 + 27

...

9 x 223 + 90

9 x 223 + 99

9 x 223 + 108

9 x 223 + 117

...

9 x 223 + 198

9 x 223 + 207

9 x 223 + 216

9 x 223 + 225

We can see that every 9th term (starting with the 9th term) has a digit sum of 9, and the difference between the digit sum and 9 is always the same (e.g. the digit sum of 45 is (4+5) = 9, and the difference is 0). Therefore, we can use the formula above to find the 2014th term:

n = 223 x 9 + ((2014 - (223 x 9)) / 9)

d = (n - 1) - 9 x ((n - 1) / 9)

Plugging in the values, we get:

n = 2011

d = 2

Therefore, the 2014th term in the sequence is:

9 x 2011 + 2 = 18099

Alternatively, if we list out the terms up to 2014, we get:

...

18027

18036

18045

18054

18063

18072

18081

18090

18099

...

So the 2014th term is 18099.

Answer 2

To solve this problem, we need to find the 2014th number in the sequence of positive integers whose digit sum is 9.

Let's observe the pattern in the sequence. Each number in the sequence is a multiple of 9 and has a digit sum of 9. The first few numbers in the sequence are:

9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 108, 117, ...

We can see that the nth number in the sequence is given by $9n$ (where $n$ starts from 1). So, to find the 2014th number in the sequence, we can simply multiply 9 by 2014:

$9 \times 2014 = 18126.$

So, the 2014th number in the sequence is 18126.