Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Hi Stoolfi,
i like your approach in solving this problem. please explain WHY you use 7 (unit digit) and raise to the 6th power (or using 6 sevens). i'm not sure the logic behind it. where do you get 6 from (the total number of digits, which is 8 in this case? or is there a special rule/formula?) thanks again. D

Well, allycat, the more I think about this problem, the less assured I am about my approach being correct. In fact, it's wrong.

I will eat crow and wipe the egg off of my face. I should know better.

The longer (and probably correct) way:

7^1 leaves remainder of 2
7^2 leaves remainder of 4
7^3 leaves remainder of 3
7^4 leaves remainder of 1
7^5 leaves remainder of 2
7^6 leaves remainder of 4
7^7 leaves remainder of 3
7^8 leaves remainder of 1

If the pattern continues, any multiple of four (other than zero) will leave a remainder of 1.

Any multiple of two that is not a multiple of four will leave a remainder of 4.

You can split 9826 in half, and wind up with an integer only once. So it is a multiple of two, but not four. So the remainder will be four.

Hey S. your new approach makes perfect sense to me. thank you.

i like the previous approach you used because it was so simple. hence, i thought you knew some shortcuts. perhaps you can tell me what you were thinking when you decided to raise 7 to the 6th power -- which i think may work as well 7^6 (raise the unit digit by the unit digit of the exponent)

Hallo guys, i came up with something but i need your opinion:
((9825+1)+1)^9826 so it seems to me that the remainder should be 2, since 9825 is exactly divisible by 5,please give me your advise

stoolfi,
Thanks. You approach is perfectly fine too. In fact, mine is very similiar to yours.
------

Here's how I did it:

You only have to focus on the unit digit, 7 in this case, of the actual number. You raise 7 to power by first few integers, i.e. 1, 2, 3, 4, 5, 6. You don't even need to multiply out whole the number, you can just get away with finding the unit digit. i.e. 7, 9, 3, 1, 7, 9, etc. You can see that the patter in the unit digit place takes place and that the patter repeats every four steps.

So, I divided 9826 by 4 and got remainder 2. That tells me that there four complete patterns plus 2 extra steps. Now I know that raising 7 to 9824 has to leave 1 as the unit digit. I just go two steps further and I get 9 in the unit digit. Just divide 9 by 5 and you get remainder 4!

I don't know if you care for the timing but I could solve this problem in 50 sec using this trick. It could be done faster but I wanted to double check my answer. The trick is pretty straight forward, I suppose.

7*7 is 49 keep the 9, *7 --63, keep the 3, *7 --21 keep the 1, *7 --7 keep the 7, *7 --49 Keep the 9, *7 --63

Any number ending in 3, divided by 5, yields a remainder of 3.

i did it this way also....but as you mentioned you used an extra 7, you should have stopped at the point when the units digit was 9 and thus if you divide anything with a units digit of nine by 5 you get a remainder of 4....so why did you switch your method stoolfi??

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...