S98-12

Yes. Jist notice that we are dividing by 100 not by 10.

Check my solution here: what-is-the-remainder-after-division-by-100-of-121666.html

Hope it helps.

Bunuel is correct to the extent of finding rem when divided by 100. even i followed the same "cyclicity" of remainders approach for ten's and unit's digit.
But for this particular problem, since all answer choices have different unit's digit, we can reduce our labor to finding just the unit's digit.

7^4 = 2401. which will leave a remainder 1 when divided with 100. Therefore, 7^4n (7^8) will also leave a remainder 1. 7^10=7^8.7^2 so the remainder will be 7^2 or 49.

Did it by approximation:
we have --> 7^10/10^2
--> we can say --> 49 x 7^8/ 10^2 --> 0.49 x 7^8 --> we know the number for 7^ 8 will end in 1. So shift decimal two spaces left , which is 1, then 3, leaving us with 9. So the product will have number ending 9 in it. Answer = 49

Madhavi1990
How did you arrive at 7^8 will end in 1??

Hello AA2014,

Please note that the approach should be decided on the basis of the divisibility test.

100 = 25 * 4.
For both 25 and 4, the divisibility test says that we check the last 2 digits of the number given, to ascertain whether the last 2 digits form a number that is divisible by 25 OR 4.
Therefore, even for 100, the divisibility test is to check for the last 2 digits of the given number. This tells us that we need to focus on finding the last 2 digits of the given number \(7^{10}\).

Does it actually simplify things or complicate it further? It actually simplifies things. But how, you may ask?

In the units digit cycle, we find out the first few units digit by multiplying the respective units digit. In case of last 2 digits also, the process is somewhat similar.

“To find out the last 2 digits, focus on finding the product of the last 2 digits of the respective numbers. The last 2 digits of this product will give you the desired answer”.

With this in mind, let’s look at establishing a pattern.

\(7^2\) = 49.

To find the last 2 digits of \(7^4\), we multiply 49 with 49 and find that the last 2 digits are 01. We don’t have to worry about the remaining digits of the result.
If \(7^4\) ends with 01, \(7^8\) =\( 7^4 * 7^4\) and hence last two digits will be the product of 01 * 01 which is clearly 01 again.

\(7^{10}\) = \(7^8\) * \(7^2\) = _ _ _ _ _ 01 * 49 = _ _ _ _ _ 49.
The last 2 digits of \(7^{10}\) is 49. Therefore, when \(7^{10}\) is divided by 100, the remainder will be 49.

Summarising, you cannot just decide the remainder based on the units digit alone. One may get away scot free in this question because the options are easy, but if there are two options ending with 9, then you cannot just decide the remainder based on the units digit.

Hope that helps!

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