# Math Help - Fun and Games

1. ## Fun and Games

A caterpillar every day in order to double the rate of growth, with 30 days to 20 cm, what spent days when it grows to 5 cm?

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2. ## Re: Fun and Games

That's impossible to answer without knowing the original length.

3. ## Re: Fun and Games

Hello, cobooboc!

I assume your native language is not English.

A caterpillar doubles its length every day and grows to 20 cm in 30 days.
How many days does it take to grow to 5 cm?

With a little Common Sense, a child could solve this problem.

The caterpillar doubles its length every day.

On the 30th day, it is 20 cm long.

Hence, on the 29th day, it was 10 cm long.

Therefore, on the 28th day, it was 5 cm long.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If an algebraic solution is required, here it is . . .

The growth is a geometric sequence with first term $a$
. . and common ratio $r = 2.$

Its length on the $n^{th}$ day is: . $a_n \:=\:a\cdot 2^{n-1}$

We are told that: $a_{30} \,=\,20$
So we have: . $a\cdot 2^{29} \:=\:20 \quad\Rightarrow\quad a \:=\:\frac{20}{2^{29}}$

The function is: . $a_n \;=\;\left(\frac{20}{2^{29}}\right)\cdot 2^{n-1}$

The length is 5: . $\left(\frac{20}{2^{29}}\right)\cdot 2^{n-1} \:=\:5$

. . . . . . . . . . . . . $2^{n-1} \:=\:5\cdot\frac{2^{29}}{20} \;=\;\frac{2^{29}}{4} \;=\;\frac{2^{29}}{2^2}$

. . . . . . . . . . . . . $2^{n-1} \:=\:2^{27} \quad\Rightarrow\quad n-1 \:=\:27 \quad\Rightarrow\quad n \:=\:28$

4. ## Re: Fun and Games

Originally Posted by Soroban
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With a little Common Sense, a child could solve this problem.

The caterpillar doubles its length every day.

On the 30th day, it is 20 cm long.

Hence, on the 29th day, it was 10 cm long.

Therefore, on the 28th day, it was 5 cm long.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If an algebraic solution is required, here it is . . .

The growth is a geometric sequence with first term $a$
. . and common ratio $r = 2.$

Its length on the $n^{th}$ day is: . $a_n \:=\:a\cdot 2^{n-1}$

We are told that: $a_{30} \,=\,20$
So we have: . $a\cdot 2^{29} \:=\:20 \quad\Rightarrow\quad a \:=\:\frac{20}{2^{29}}$

The function is: . $a_n \;=\;\left(\frac{20}{2^{29}}\right)\cdot 2^{n-1}$

The length is 5: . $\left(\frac{20}{2^{29}}\right)\cdot 2^{n-1} \:=\:5$

. . . . . . . . . . . . . $2^{n-1} \:=\:5\cdot\frac{2^{29}}{20} \;=\;\frac{2^{29}}{4} \;=\;\frac{2^{29}}{2^2}$

. . . . . . . . . . . . . $2^{n-1} \:=\:2^{27} \quad\Rightarrow\quad n-1 \:=\:27 \quad\Rightarrow\quad n \:=\:28$
yes,You can turn count, it will be relatively simple, but your algorithm is correct, and I admire you.

5. ## Re: Fun and Games

Originally Posted by HallsofIvy
That's impossible to answer without knowing the original length.
You think is wrong, in fact, you can calculate its original length, you go and try.

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6. ## Re: Fun and Games

My mistake. Now I see that your post was non-sense only intended to allow you to link to a commercial site- in other words, spam.

7. ## Re: Fun and Games

Originally Posted by HallsofIvy
My mistake. Now I see that your post was non-sense only intended to allow you to link to a commercial site- in other words, spam.
If you do not, you can completely ignore it and hope you do not misunderstand, I am a mathematics graduate math or feelings.