Events & Promotions
|
|
GMAT Club Daily Prep
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2301:30 AM EDT
-02:30 AM EDT
Struggling with GMAT Verbal as a non-native speaker? Harsh improved his score from 595 to 695 in just 45 days—and scored a 99 %ile in Verbal (V88)! Learn how smart strategy, clarity, and guided prep helped him gain 100 points.
Apr 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Apr 2212:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
22 Jan 2020, 02:11
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (01:57) correct
29%
(02:00)
wrong
based on 607
sessions
History
Date
Time
Result
Not Attempted Yet
What is the unit's digit in the numerical value of \((2)^{37} * (3)^{47} * (7)^{79} * (13)^{26}\)?
A. 2
B. 4
C. 6
D. 8
E. 9
A. 2
B. 4
C. 6
D. 8
E. 9
22 Jan 2020, 02:30
Kudos
Bookmarks
Bunuel
Cyclicity of Unit digit of 2 = 4 (i.e. Unit digit of powers of 2 repeat after every 4th power)
e.g.
\(2^1 = 2\)
\(2^2 = 4\)
\(2^3 = 8\)
\(2^4 = 16\)
\(2^5 = 32\)
\(2^6 = 64\)
\(2^7 = 128\)
\(2^8 = 256\)
\(2^9 = 512\)
Similarly
Cyclicity of Unit digit of 3 = 4
Cyclicity of Unit digit of 7 = 4
Cyclicity of Unit digit of 13 = Cyclicity of Unit digit of 3 = 4 (For unit digit calculation, only the unit digit of question matters)
i.e. Now we will divide the power of each of these numbers with their cyclicity and the remainder will be used teh final power of number responsible to render the unit digit of teh number
Unit digit of \((2)^{37}\) = Unit digit of \((2)^{36+1}\) = Unit digit of \((2)^{1}\) = 2
Unit digit of \((3)^{47}\) = Unit digit of \((3)^{44+3}\) = Unit digit of \((3)^{3}\) = 7
Unit digit of \((7)^{79}\) = Unit digit of \((7)^{76+3}\) = Unit digit of \((7)^{3}\) = 3
Unit digit of \((13)^{26}\) = Unit digit of \((3)^{26}\) = Unit digit of \((3)^{24+2}\) = Unit digit of \((3)^{2}\) = 9
i.e. Unit digit of \((2)^{37} * (3)^{47} * (7)^{79} * (13)^{26}\) = Unit digit of \(2*7*3*9\) = 8
Answer: Option D
28 Jan 2020, 23:50
Kudos
Bookmarks
This is a very simple and direct question on finding the units digit. Note that although the expression given looks humongous and terrifying, it is actually not so. That’s because if you know the cyclicity of unit digits, then you only need to focus on the exponents and completely disregard the magnitude of the bases.
Remember, I said magnitude of the bases. But you would still need to consider the unit digit of the bases because these will tell you which cycle to consider.
The cyclicity of all the numbers given in the base here i.e. 2, 7, 3 and 13 is 4. This essentially means that the units digit of different powers of these numbers repeat in intervals of 4.
For example, let us take the powers of 2.
\(2^1\) = 2, \(2^2\) = 4, \(2^3\) = 8, \(2^4\) = 16, \(2^5\) = 32, \(2^6\) = 64 and so on. You see that the unit digit of these numbers is repeating in intervals of 4. This frequency of repetition is called cyclicity.
If any number ends with 2 or 3 or 7 or 8, the cyclicity of the units digit of different powers of that number will be 4. This means, if we make columns of the units digit of different powers, we will have 4 different columns.
To quickly find out the units digit of the power of a number, divide the power by its unit digit cyclicity. If the remainder is 1, you can find your unit digit in the first column; if the remainder is 2, unit digit is in the second column and so on.
If the remainder is ZERO, it means that the power/exponent is a multiple of 4 i.e. the cyclicity and hence the units digit can be found in the 4th column.
\(2^{37}\). 37 divided by 4 leaves a remainder of 1. The cycle of units digit of numbers ending with 2 is 2 – 4 – 8 – 6. Since the remainder is 1, the unit digit is from the first column i.e. 2.
So, \(2^{37}\) is a big number ending with 2.
\(3^{47}\). 47 divided by 4 leaves a remainder of 3. The cycle of units digit of numbers ending with 3 is 3 – 9 – 7 – 1. Since the remainder is 3, the unit digit is from the third column i.e. 7.
So, \(3^{47}\) is a big number ending with 7.
\(7^{79}\). 79 divided by 4 leaves a remainder of 3. The cycle of units digit of numbers ending with 7 is 7 – 9 – 3 – 1. Since the remainder is 3, the unit digit is from the third column i.e. 3.
So, \(7^{79}\) is a big number ending with 3.
\(13^{26}\). 26 divided by 4 leaves a remainder of 2. The cycle of units digit of numbers ending with 3 is 3 – 9 – 7 – 1. Since the remainder is 2, the unit digit is from the second column i.e. 9.
So, \(13^{26}\) is a big number ending with 9.
Now, to find the final units digit of the entire expression, we multiply the units digit of the respective numbers i.e. 2*7*3*9 which gives us a number ending with 8. The units digit of the expression is 8.
The correct answer option is D.
Hope that helps!
Remember, I said magnitude of the bases. But you would still need to consider the unit digit of the bases because these will tell you which cycle to consider.
The cyclicity of all the numbers given in the base here i.e. 2, 7, 3 and 13 is 4. This essentially means that the units digit of different powers of these numbers repeat in intervals of 4.
For example, let us take the powers of 2.
\(2^1\) = 2, \(2^2\) = 4, \(2^3\) = 8, \(2^4\) = 16, \(2^5\) = 32, \(2^6\) = 64 and so on. You see that the unit digit of these numbers is repeating in intervals of 4. This frequency of repetition is called cyclicity.
If any number ends with 2 or 3 or 7 or 8, the cyclicity of the units digit of different powers of that number will be 4. This means, if we make columns of the units digit of different powers, we will have 4 different columns.
To quickly find out the units digit of the power of a number, divide the power by its unit digit cyclicity. If the remainder is 1, you can find your unit digit in the first column; if the remainder is 2, unit digit is in the second column and so on.
If the remainder is ZERO, it means that the power/exponent is a multiple of 4 i.e. the cyclicity and hence the units digit can be found in the 4th column.
\(2^{37}\). 37 divided by 4 leaves a remainder of 1. The cycle of units digit of numbers ending with 2 is 2 – 4 – 8 – 6. Since the remainder is 1, the unit digit is from the first column i.e. 2.
So, \(2^{37}\) is a big number ending with 2.
\(3^{47}\). 47 divided by 4 leaves a remainder of 3. The cycle of units digit of numbers ending with 3 is 3 – 9 – 7 – 1. Since the remainder is 3, the unit digit is from the third column i.e. 7.
So, \(3^{47}\) is a big number ending with 7.
\(7^{79}\). 79 divided by 4 leaves a remainder of 3. The cycle of units digit of numbers ending with 7 is 7 – 9 – 3 – 1. Since the remainder is 3, the unit digit is from the third column i.e. 3.
So, \(7^{79}\) is a big number ending with 3.
\(13^{26}\). 26 divided by 4 leaves a remainder of 2. The cycle of units digit of numbers ending with 3 is 3 – 9 – 7 – 1. Since the remainder is 2, the unit digit is from the second column i.e. 9.
So, \(13^{26}\) is a big number ending with 9.
Now, to find the final units digit of the entire expression, we multiply the units digit of the respective numbers i.e. 2*7*3*9 which gives us a number ending with 8. The units digit of the expression is 8.
The correct answer option is D.
Hope that helps!
