A lot of 5's ...

The first step here was to look for a pattern.

$5 \times 55 \mod 100 = 75 \\ 5 \times 55 \times 555 \mod 100 = 25 \\ 5 \times 55 \times 555 \times 5555 \mod 100 = 75 \\ 5 \times 55 \times 555 \times 5555 \times 5555 \mod 100 = 25$

And so on.. now, observe that this gives us a rough conditional (I am a programmer, so..):

if numOf5sInLastNumber mod 2 is 0, then answer is 75, else 25.

Putting in 5555, which is not a multiple of 2, I get 25 which is the answer.

5 55%100=25, then 25 55%100=75, and 75*55%100=25, So if there's an even number of times, it's 75, and if it's an odd number of times, it's 25. Since 5555 is odd, the answer is 25

Take mod 100 each term in the product to see how it will affect the remainder after being divided by 100. The question becomes: What is the remainder when the following is divided by 100?: $5 \times 55 \times 55 \times 55 .... \times 55$ with 5554 55's.

Next, I tried multiplying the first few terms out (remembering to take mod 100 of each term) and seeing if a pattern could be determined.
1st Term: $5$
2nd Term: $5 \times 55 = 275= 75 mod 100.$
3rd Term: $75 \times 55 = 75 \times 52 + 3 \times 75 = 225 mod 100 = 25 mod 100.$
Note: 52 times 75 becomes 0 after taking mod 100, since 4 times any integer times 75 is going to be a multiple of 300.
4th Term: $25 \times 55 = 52 \times 25 + 3 \times 25 = 0+ 75 = 75 mod 100$ by the same logic above.
There seems to be a pattern: After multiplying 25 by 55, a remainder of 75 is left over and after multiplying 75 by 55, a remainder of 25 is left over.
Now, It can be said that after an odd number of terms are multiplied together (in all cases except the first term), the remainder is 25; every even number of terms product ends has a remainder of 75.

Now, since 5555 is an odd number, it can be said that the remainder after being divided by 100 is $\boxed{25}.$

$\begin{aligned} &~ &5\times 55\times 555\times \cdot\cdot\cdot\times \underbrace{555...555}_{5555~ \text{fives}} \nonumber \\ \\ &\equiv &5\times \underbrace{55\times 55\times 55 \times \cdot\cdot\cdot\times 55}_{5554~\text{fifty fives}} \pmod{100} \nonumber \\ \\ &=& 5\times 55^{5554} \nonumber \\ \\ &\equiv & 5\times 25 \pmod{100} \nonumber \\ \\ &=&125 \nonumber \\ \\ &\equiv &\fbox{25} \pmod{100} \end{aligned}$

Using this pattern for 5 55 = 275 for 5 55 555=152625 for 5 55 555 5555=8479081875 for the last factor having even "5"s, the remainder when divided by 100 is 75 for the last factor having odd"5 "s, the remainder when divided by 100 is 25

for the problem, since the last factor is odd, then the answer must be 25

We can solve this problem by looking for the pattern .

The remainder of a number divided by 100 is same to the last two digits of the number.Therefore we need to find the last two digits of $5\times55\times555\times...\times555...55$ } $5555$ .

(Note: $555...55$ } $5555$ means there are $5555$ digits in that number.)

Now, let's look for the pattern:

$5\times55=275$ Last two digits: $75$

$5\times55\times555=152625$ Last two digits: $25$

$5\times55\times...\times5555=847831875$ Last two digits: $75$

$5\times...\times55555=47101299815625$ Last two digits: $25$ ......

Can you see the pattern?The pattern is the last two digit of the products have the sequence: $75,25,75,25...$

We can see that the product ends with the digits $75$ is a product of an even number of numbers(in the problem).For example, $5\times55=275$ .The product ends with digits $75$ ,and it is a product of two numbers, $5$ and $55$ .

On the other hand, the product ends with digits $25$ is a product of an odd number of numbers(in the problem).By this pattern,

$5\times55\times555\times...\times555...55$ } $5555\rightarrow5555$ numbers.( odd number )

Therefore,the remainder after dividing $100$ (the last two digits) of the product is $\boxed{25}$ .

.

First we realize that each term except for the first in $5 \times 55 \times 555 \times ... \times 555..55$ , where the last term is $5555$ numbers of $5'$ s, are all equal to $55 mod(100)$ due to splitting up the number into two parts, one consisting of trailing $55'$ s and the other of the $5'$ s before that.

Constructing a table yields: $n$ is the number of \(55')s

1. \(n=1, 55 mod(100)\).

1. $n=2, 25 mod(100)$ .

2. $n=3, 75 mod(100)$ .

3. $n=4, 25 mod(100)$ .

At $n=4$ we stop as we already know that the next element of the list will be $75$ . We can thus conclude that as there are $5554$ of terms that are equal to $55 mod(100)$ , the propduct of all these $5554$ terms is equal to $25 mod(100)$ as $5554$ is even. Finally, we are left with a $5$ , the answer can then be calculated: $5 \times 25=125=25 mod(100)$ . Answer: $\boxed{25}$

This can be solved with a little smaller-scale experimentation. To start, the remainder when a number is divided by 100 is the last 2 digits. So, 5 55 have last two digits 75, multiplying this by 555 gives last two 25, 5555 gives 75. and so on. From this, it is clear that there is an alternating pattern within the remainders. Since the last number (i.e. 5555) has an odd number digits, we look for when we last multiplied an odd number of digits was, which was 5 55*555, which gave us a remainder of 25, which is our answer.

Here we make an observation that if 5 is multiplied by 55 we get 275 which on dividing by 100 gets remainder 25 .Similarly if 5 is 1st multiplied by 55 and then by 555 and dividing the result with 100 we get remainder 75 . Hence, we can say that when 5 is multiplied by 555555...555555(n times) where n is even we always get remainder 25

Based on the properties of modulo:

$5 = 5 (mod 100)$

$55 = 55 (mod 100)$

$555 = 55 (mod 100)$

and so on.

Therefore the remainder of $5 \times 55 \times 555 \times ... \times 555...55$ given that the last term of the product has $5555$ digits is $(5 \times 55 \times 55 \times ... \times 55)(mod 100)$

$(5 \times 55 \times 55 \times ... \times 55)(mod 100)$ can have two values depending on the number of terms given that the number of terms is greater than $1$

$(5 \times 55 \times 55 \times ... \times 55)(mod 100) = 25(mod 100)$ if the number of terms is odd and

$(5 \times 55 \times 55 \times ... \times 55)(mod 100) = 75(mod 100)$ if the number of terms is even.

The last term has 5555 digits made out of 5's. Therefore, the last number is the $5555^{th}$ term, this makes the number of terms 5555. Based on the conditions given, the remainder is 25

when divided by 100 when number of 5's is odd then remainder is 75,

5 × 55 total 5's = 3

= 275 remainder is 75 when number of 5's even then remainder is 25

5 × 55 × 555 total 5's = 6

= 152625 remainder is 25

as there are 5555 in last term total number of 5's = 5555 × 5556/2 = 5555 × 2778 is even

so remainder is 25

Well... Firstly, try to think like this- 5X55 and the answer is 275 and reminder- 75; 5X55X555 and the answer is 152625 and reminder- 25; 5X55X555X5555 and the .... and reminder- 75; again, 5X55X555X5555X55555 and the .... and reminder- 25;

So we can see that, if the biggest number's number of digits are even then, the reminder is 75 and if it is odd, then the reminder is 25.

So, in the question biggest number has 5555 digits and it is odd.

So the reminder is 25 :D :D

Original solution I guess:

First, we know that the rest when divided by 100 means the 2 last digit of the number. Let's write it differently:

$5^{5555} \times (1 \times 11 \times 111 \times . . . \times 111....111)$

Well, what are the last 2 digits of $5^{5555}$ ? It's 25 of course, you can see this pattern on ( 5, 125, 625, 3125....) (because $25 \times 5$ is 125, the 2 last digits won't change)

So we have (a number ending with 25) $\times ( 1 \times 11 \times 111 \times . . . \times 111....111)$

Now, what happens to the last 2 digits when you multiply a number by 11? Let's call the 2 last digits ( 2 and 5) as x and y. When multiplied by 11:

(. . . . . .) x y

(. . . .) x y (this corresponds to the sum you do on the paper after multiplying, (y + 0, x + y, ....))

Last digit (y) maintains and the other one (x) gets incremented by y

You can observe that the same thing happens with 111, 1111, and so on, for the last 2 digits

So we know already that the last digit of this multiplication is 5 The other one would be 2 plus 5 for each multiplication, being ( 7, 2, 7, 2, 7 ....)

We have 5554 multiplications ( 11 to 111....11 (5555 "one") ), so the digit will be 2

25 is the answer

5 55/100 = R-25,5 55*555/100=R- 75 PATTERN = 5,25,75,25,75 SO ANSWER = 25