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 500,000 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:

Lets check how to drill this. We will discuss the last two digits of numbers ending with the following digits in sets: a) 1 b) 3,7&9 c) 2, 4,6&8 d) 5

a) Number ending with 1:

Ex : Find the last 2 digits of 31^786 Now, multiply the 10s digit of the number with the last digit of exponent 31^786=3*6=18 ==> 8 is the 10s digit. Units digit is obviously 1 So, last 2 digits are => 81

b) Number ending with 3 ,7 & 9:

Ex: Find last 2 digits of 19^266 We need to get this in such as way that the base has last digit as 1 19^266 = (19^2)^133 = 361^133 Now, follow the previous method => 6 * 3 = 18 = 18 8 is 10s digit. So, last two digits are => 81

Remember: 3^4=81 7^4=2401 9^2 = 81

Ex 2: Find last two digits of 33^288 Now, 33^288 = (33^4)^72 = (xx21 )^72 Tens digit is -> 2*2 = 04 -> 4 So, last two digits are => 41

Ex 3: find last 2 digits of 87^474 (87^2)*(87^4)^118 => (xx69) * (xx61 )^118 ==> (6 x 8 = 48 ==> 8 is 10s digit) => (xx69)*(81) So, last two digits are 89

c)Ending with 2, 4, 6 or 8:

Here, we use the fact that 76 power any number gives 76. i.e 76^n = ...xxx76 (e.g. 76^2 = 5,776, 76^3 = 438,976) We also need to remember that, 24^2 = xx76 2^10 xx24 24^even = xx76 24^odd = xx24

Ex: Find the last two digits of 2^543 2^543 = ((2^10)^54) * (2^3) = ((xx24)54)* 8 = ((xx76)^27)*8 76 power any number is 76 Which gives last digits as => 76 * 8 = 608 So last two digits are : 08

c)Ending with 5:

5 has its own special cases: The tens digit of a number raised to a power changes based on tens digits of the number itself. Say, we have following number: 05, 15, 25, 35…

say--> x = 05 x^2 = 25 x^3 = 125 x^4 = 625 as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 15 x^2 = 225 x^3 = 3375 x^4 = 50625 x^5 = 759375 as the tens digit of x is 1(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on.

x = 25 x^2 = 625 x^3 = ..XX25 x^4 = ..XX25 x^5 = ..XX25 as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 35 x^2 = 1225 x^3 = ..XX75 x^4 = ..XX25 x^5 = ..XX75 as the tens digit of x is 3(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on

From above, we see that (1) if number's 10s digit is even, then last digit will always has to be 25. (2) if number's 10s digit is odd, then last digit will always has to be in cyclic pattern of 25, 75,25,75 and so on...

Now, again, try the question posted on the top.
_________________

Thanks and Regards!

P.S. +Kudos Please! in case you like my post.

Last edited by goutamread on 03 Jan 2013, 10:17, edited 1 time in total.

Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]

Show Tags

03 Jan 2013, 09:07

Very laborious isnt it?
_________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]

Show Tags

03 Jan 2013, 09:19

2

This post received KUDOS

Question:The last two digits of (31^786)(19^266)(2^101) is A. 14 B. 22 C. 36 D. 72 E. 92

Solution: 31^786==> Unit Digits: as last digit is 1, units digit has to be 1. Tens digit: 31^786 ==> as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=18 and thus, 8 is the tens digit. Thus, last two digit of expression 31^786 = 81

19^266 ==> Unit Digits: as last digit is 9, as we know pattern power raise to 9: 9,1,9,1... i.e. it repeats after every 2 iterations.. as 266 is even i.e. dividible by 2, we cud say,units digit has to be 1. Tens digit: 19^266 = (19^2)^133 = (xx61)^133 ==> now, as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=18 and thus, 8 is the tens digit. Thus, last two digit of expression 19^266 = 81

(2^101) ==> Tens Digit: 2^101 = 2^100 * 2 =[(2^10)^10 ]*2 = [XX24^10]*2 ==> As we know xx24^even = xx76 ==> [xx76 ^ 10]*2 ==> as we knoe anything raised to xx76 is xx76 ==> xx76*2 = XX52

Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]

Show Tags

03 Jan 2013, 09:27

1

This post received KUDOS

rajathpanta wrote:

Very laborious isnt it?

Rajath - This example was just to elaborate all possible scenarios i.e. (a)1 (b)3,7,9 and (c)2,4,6,8. Indeed, I have detailed each step over here. Once you are used to the concept/rules, this problem shall not take more than the (per question) average time.
_________________

Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]

Show Tags

03 Jan 2013, 09:55

[quote="goutamread"]Question:The last two digits of (31^786)(19^266)(2^101) is A. 14 B. 22 C. 36 D. 72 E. 92

Solution: 31^786==> Unit Digits: as last digit is 1, units digit has to be 1. Tens digit: 31^786 ==> as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=18 and thus, 8 is the tens digit. Thus, last two digit of expression 31^786 = 81

19^266 ==> Unit Digits: as last digit is 9, as we know pattern power raise to 9: 9,1,9,1... i.e. it repeats after every 2 iterations.. as 266 is even i.e. dividible by 2, we cud say,units digit has to be 1. Tens digit: 19^266 = (19^2)^133 = (xx61)^133 ==> now, as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=18 and thus, 8 is the tens digit. Thus, last two digit of expression 19^266 = 81

(2^101) ==> Tens Digit: 2^101 = 2^100 * 2 =[(2^10)^10 ]*2 = [XX24^10]*2 ==> As we know xx24^even = xx76 ==> [xx76 ^ 10]*2 ==> as we knoe anything raised to xx76 is xx76 ==> xx76*2 = XX52

THUS XX81 * XX81 *XX52 = XX72[/quotew

what would be second last digit if last digit is 5

Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]

Show Tags

03 Jan 2013, 10:16

1

This post received KUDOS

chiknichameleon wrote:

what would be second last digit if last digit is 5

Actually, 5 has its own special cases: The tens digit of a number raised to a power changes based on tens digits of the number itself. Say, we have following number: 05, 15, 25, 35…

say--> x = 05 x^2 = 25 x^3 = 125 x^4 = 625 as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 15 x^2 = 225 x^3 = 3375 x^4 = 50625 x^5 = 759375 as the tens digit of x is 1(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on.

x = 25 x^2 = 625 x^3 = ..XX25 x^4 = ..XX25 x^5 = ..XX25 as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 35 x^2 = 1225 x^3 = ..XX75 x^4 = ..XX25 x^5 = ..XX75 as the tens digit of x is 3(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on

From above, we see that (1) if number's 10s digit is even, then last digit will always has to be 25. (2) if number's 10s digit is odd, then last digit will always has to be in cyclic pattern of 25, 75,25,75 and so on...

P.S. Thanks for asking the question. I missed that part. I shall update the post to include that part.
_________________

Lets check how to drill this. We will discuss the last two digits of numbers ending with the following digits in sets: a) 1 b) 3,7&9 c) 2, 4,6&8 d) 5

a) Number ending with 1:

Ex : Find the last 2 digits of 31^786 Now, multiply the 10s digit of the number with the last digit of exponent 31^786=3*6=18 ==> 8 is the 10s digit. Units digit is obviously 1 So, last 2 digits are => 81

b) Number ending with 3 ,7 & 9:

Ex: Find last 2 digits of 19^266 We need to get this in such as way that the base has last digit as 1 19^266 = (19^2)^133 = 361^133 Now, follow the previous method => 6 * 3 = 18 = 18 8 is 10s digit. So, last two digits are => 81

Remember: 3^4=81 7^4=2401 9^2 = 81

Ex 2: Find last two digits of 33^288 Now, 33^288 = (33^4)^72 = (xx21 )^72 Tens digit is -> 2*2 = 04 -> 4 So, last two digits are => 41

Ex 3: find last 2 digits of 87^474 (87^2)*(87^4)^118 => (xx69) * (xx61 )^118 ==> (6 x 8 = 48 ==> 8 is 10s digit) => (xx69)*(81) So, last two digits are 89

c)Ending with 2, 4, 6 or 8:

Here, we use the fact that 76 power any number gives 76. i.e 76^n = ...xxx76 (e.g. 76^2 = 5,776, 76^3 = 438,976) We also need to remember that, 24^2 = xx76 2^10 xx24 24^even = xx76 24^odd = xx24

Ex: Find the last two digits of 2^543 2^543 = ((2^10)^54) * (2^3) = ((xx24)54)* 8 = ((xx76)^27)*8 76 power any number is 76 Which gives last digits as => 76 * 8 = 608 So last two digits are : 08

c)Ending with 5:

5 has its own special cases: The tens digit of a number raised to a power changes based on tens digits of the number itself. Say, we have following number: 05, 15, 25, 35…

say--> x = 05 x^2 = 25 x^3 = 125 x^4 = 625 as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 15 x^2 = 225 x^3 = 3375 x^4 = 50625 x^5 = 759375 as the tens digit of x is 1(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on.

x = 25 x^2 = 625 x^3 = ..XX25 x^4 = ..XX25 x^5 = ..XX25 as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 35 x^2 = 1225 x^3 = ..XX75 x^4 = ..XX25 x^5 = ..XX75 as the tens digit of x is 3(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on

From above, we see that (1) if number's 10s digit is even, then last digit will always has to be 25. (2) if number's 10s digit is odd, then last digit will always has to be in cyclic pattern of 25, 75,25,75 and so on...

Now, again, try the question posted on the top.

The answer is 72 31^786 = (30+1)^786 use binomial theorem The last two digits will be 786*30 + 1 = 81 19^266 = 61^133 133*60+1 = 81 2^101=(2^10)^10*2=24^10*2=76^5*2=76*2=52(the last two digits of any power of 76 is always 76) 81*81*52=61*52=72

gmatclubot

Re: The last two digits of (31^786)(19^266)(2^101) is
[#permalink]
10 Apr 2016, 05:33

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...