cody has 22 rare coins. each year cody plans to add 15 coins to his collection. How many years will it take Cody to collect 577 coins?
To figure out how many years it will take Cody to collect 577 coins, we can set up an equation.
Let's assume it takes Cody "y" number of years to collect 577 coins. In each year, Cody adds 15 coins to his collection.
So, in the first year, Cody will have a total of 22 + 15 coins = 37 coins.
In the second year, Cody will have 37 + 15 coins = 52 coins.
Similarly, in the third year, Cody will have 52 + 15 coins = 67 coins.
We can observe that the number of coins Cody has each year follows a pattern: 22, 37, 52, 67, ...
This can be represented as an arithmetic sequence, where the first term (a₁) is 22, and the common difference (d) is 15.
The nth term (aₙ) of an arithmetic sequence can be calculated using the formula: aₙ = a₁ + (n - 1) × d.
So, aₙ = 22 + (n - 1) × 15.
Now, we need to find which value of 'n' will make aₙ equal to or greater than 577.
22 + (n - 1) × 15 ≥ 577.
Simplifying this equation, we get:
15n - 15 ≥ 555.
Adding 15 to both sides, we have:
15n ≥ 570.
Dividing both sides by 15, we get:
n ≥ 38.
Since 'n' represents the number of years Cody needs to collect 577 coins, Cody will be able to collect 577 coins in 38 years or more.
Therefore, it will take Cody at least 38 years to collect 577 coins.
To find out how many years it will take Cody to collect 577 coins, we can use basic algebra.
Let's assume that Cody takes "x" years to collect 577 coins.
In the first year, Cody starts with 22 coins and adds 15 coins. So after the first year, Cody will have a total of 22 + 15 = 37 coins.
In the second year, Cody adds another 15 coins to his collection. So after the second year, he will have a total of 37 + 15 = 52 coins.
If we continue this pattern, we can see that Cody's total number of coins after each year forms an arithmetic sequence, where the common difference is 15.
We can write the arithmetic sequence as: 22, 37, 52, ...
To find the number of years it takes for Cody to collect 577 coins, we need to find the nth term of the arithmetic sequence that is equal to or greater than 577.
We can use the arithmetic sequence formula:
an = a1 + (n - 1)d
where:
an = nth term of the sequence
a1 = first term of the sequence
d = common difference
In this case, we want to find the smallest value of n such that an ≥ 577.
Let's plug in the values in the formula:
577 = 22 + (n - 1)15
Simplifying the equation:
577 - 22 = 15n - 15
555 = 15n - 15
570 = 15n
n = 570/15
n = 38
Therefore, it will take Cody 38 years to collect 577 coins.
In the first year, Cody will have 22 + 15 = <<22+15=37>>37 coins.
In the second year, he will have 37 + 15 = <<37+15=52>>52 coins.
In the third year, he will have 52 + 15 = <<52+15=67>>67 coins.
In the fourth year, he will have 67 + 15 = <<67+15=82>>82 coins.
In the fifth year, he will have 82 + 15 = <<82+15=97>>97 coins.
In the sixth year, he will have 97 + 15 = <<97+15=112>>112 coins.
In the seventh year, he will have 112 + 15 = <<112+15=127>>127 coins.
In the eighth year, he will have 127 + 15 = <<127+15=142>>142 coins.
In the ninth year, he will have 142 + 15 = <<142+15=157>>157 coins.
In the tenth year, he will have 157 + 15 = <<157+15=172>>172 coins.
In the eleventh year, he will have 172 + 15 = <<172+15=187>>187 coins.
In the twelfth year, he will have 187 + 15 = <<187+15=202>>202 coins.
In the thirteenth year, he will have 202 + 15 = <<202+15=217>>217 coins.
In the fourteenth year, he will have 217 + 15 = <<217+15=232>>232 coins.
In the fifteenth year, he will have 232 + 15 = <<232+15=247>>247 coins.
In the sixteenth year, he will have 247 + 15 = <<247+15=262>>262 coins.
In the seventeenth year, he will have 262 + 15 = <<262+15=277>>277 coins.
In the eighteenth year, he will have 277 + 15 = <<277+15=292>>292 coins.
In the nineteenth year, he will have 292 + 15 = <<292+15=307>>307 coins.
In the twentieth year, he will have 307 + 15 = <<307+15=322>>322 coins.
In the twenty-first year, he will have 322 + 15 = <<322+15=337>>337 coins.
In the twenty-second year, he will have 337 + 15 = <<337+15=352>>352 coins.
In the twenty-third year, he will have 352 + 15 = <<352+15=367>>367 coins.
In the twenty-fourth year, he will have 367 + 15 = <<367+15=382>>382 coins.
In the twenty-fifth year, he will have 382 + 15 = <<382+15=397>>397 coins.
In the twenty-sixth year, he will have 397 + 15 = <<397+15=412>>412 coins.
In the twenty-seventh year, he will have 412 + 15 = <<412+15=427>>427 coins.
In the twenty-eighth year, he will have 427 + 15 = <<427+15=442>>442 coins.
In the twenty-ninth year, he will have 442 + 15 = <<442+15=457>>457 coins.
In the thirtieth year, he will have 457 + 15 = <<457+15=472>>472 coins.
In the thirty-first year, he will have 472 + 15 = <<472+15=487>>487 coins.
In the thirty-second year, he will have 487 + 15 = <<487+15=502>>502 coins.
In the thirty-third year, he will have 502 + 15 = <<502+15=517>>517 coins.
In the thirty-fourth year, he will have 517 + 15 = <<517+15=532>>532 coins.
In the thirty-fifth year, he will have 532 + 15 = <<532+15=547>>547 coins.
In the thirty-sixth year, he will have 547 + 15 = <<547+15=562>>562 coins.
In the thirty-seventh year, he will have 562 + 15 = <<562+15=577>>577 coins.
Therefore, it will take Cody 37 years. Answer: \boxed{37}.