Forum Home > GMAT > Quantitative > Problem Solving (PS)
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.
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.- Dec 15
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!- Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options.
06 Oct 2020, 03:33
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
17% (03:36) correct
83%
(02:47)
wrong
based on 53
sessions
History
Date
Time
Result
Not Attempted Yet
Let n be a positive integer that has the same digit appearing in all of its place values. What is the number of digits in the smallest possible value of n such that it is divisible by all positive digits from 1 to 9, except for 5?
A. 3
B. 6
C. 9
D. 12
E. 18
A. 3
B. 6
C. 9
D. 12
E. 18
06 Oct 2020, 08:23
Kudos
Bookmarks
We have a number AAAA...AA, where all the digits are the same, and we need the number to be divisible by 1, 2, 3, 4, 6, 7, 8 and 9. If the number is divisible by 8, it will definitely be divisible by 4 and by 2. If it is divisible by 8 and 9, it will definitely be divisible by 6 and by 3. So really all we need is for our number to be divisible by 7, 8 and 9.
Now, for a number to be divisible by 8, its last three digits must make a number divisible by 8. Clearly our digit A must then be even. But the numbers 222, 444 and 666 are not multiples of 8. So the only possible value of A is 8, and we know our number looks like 8888....888.
For this number to be a multiple of 9, its digits will need to sum to a multiple of 9. That will only happen if we have 9 or 18 or 27 etc identical digits in our number. So the answer must be C or E.
To decide if C is correct, we just need to know if 888,888,888 is divisible by 7. There is some divisibility test for 7 that is completely useless for GMAT purposes (and in most areas of math in general - I have a Masters specifically in Number Theory and have literally never used that test once in my life). I imagine it would be helpful for this precise question, because if I recall correctly, it involves subtracting even and odd placed digits, which would be easy to do for this kind of number, so that divisibility test probably produces an answer instantly here. But because I don't know that test and have no reason to learn it, I'd instead just subtract large multiples of 7 from 888,888,888 until I got to a familiar number -- if we're starting from a multiple of 7, we'll need to get new smaller multiples of 7 every time we subtract a multiple of 7. So starting from 888,888,888, I subtracted 777,777,777 to get 111,111,111, then subtracted 105,000,000 to get 6,111,111 then subtracted 5,600,000 to get 511,111, then subtracted 490,000 to get 21,111, then subtracted 21,000 to get 111. That's not a multiple of 7, so 888,888,888 cannot be either. That leaves E as the only possible answer.
Now, for a number to be divisible by 8, its last three digits must make a number divisible by 8. Clearly our digit A must then be even. But the numbers 222, 444 and 666 are not multiples of 8. So the only possible value of A is 8, and we know our number looks like 8888....888.
For this number to be a multiple of 9, its digits will need to sum to a multiple of 9. That will only happen if we have 9 or 18 or 27 etc identical digits in our number. So the answer must be C or E.
To decide if C is correct, we just need to know if 888,888,888 is divisible by 7. There is some divisibility test for 7 that is completely useless for GMAT purposes (and in most areas of math in general - I have a Masters specifically in Number Theory and have literally never used that test once in my life). I imagine it would be helpful for this precise question, because if I recall correctly, it involves subtracting even and odd placed digits, which would be easy to do for this kind of number, so that divisibility test probably produces an answer instantly here. But because I don't know that test and have no reason to learn it, I'd instead just subtract large multiples of 7 from 888,888,888 until I got to a familiar number -- if we're starting from a multiple of 7, we'll need to get new smaller multiples of 7 every time we subtract a multiple of 7. So starting from 888,888,888, I subtracted 777,777,777 to get 111,111,111, then subtracted 105,000,000 to get 6,111,111 then subtracted 5,600,000 to get 511,111, then subtracted 490,000 to get 21,111, then subtracted 21,000 to get 111. That's not a multiple of 7, so 888,888,888 cannot be either. That leaves E as the only possible answer.
06 Oct 2020, 20:30
Kudos
Bookmarks
Let us eliminate the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9
Since the number should not be divisible by 5, the number cannot be 5
To be divisible by 2, 4 6 or 8, the number should be even. This eliminates the digits 1, 3, 7 and 9
To be divisible by 4, the last 2 digits should be divisible by 4. This eliminate numbers ending with 22 and 66..So the digits 2 and 6 are also eliminated.
We are left with digits 4 and 8. To be divisible by 8, the last 3 digits should be divisible by 8. 444 is not divisible by 8, so we are now left with an n digit number containing 8's and we need to ensure that the number formed is divisible by 7 and 9.
So we can have 9 8's or 18 8's
Both these numbers are divisible by 9 (sum of the digits should be divisible by 9)
To divisibility test for 7 is the number of 10's + 5 times the units place should be divisible by 7 (the number of 10's is the number up to the 10's place, for e.g. in the number 543, the number of 10's is 54. In the number 6425 it is 642)
8 8 8 8 8 8 8 8 8 = 88888888 + (5 * 8) = 88888928
Now this can be repeated and made smaller by using the same method as above = 8888892 + 8*5 = 8888932
= 888893 + (2 * 5) = 888903
= 88890 + (5 * 3) = 88905
= 8890 + (5 * 3) = 8905
= 890 + (5 * 3) = 905
= 90 + (5 * 3) = 105
= 10 + (5 * 3) = 25 and we know that 25 is not divisible by 7.
Now it is very obvious that the method is cumbersome and time consuming just to find out if the number is divisible by 7. Therefore the better method, would be to directly divide by 7.
Option E
Arun Kumar
Since the number should not be divisible by 5, the number cannot be 5
To be divisible by 2, 4 6 or 8, the number should be even. This eliminates the digits 1, 3, 7 and 9
To be divisible by 4, the last 2 digits should be divisible by 4. This eliminate numbers ending with 22 and 66..So the digits 2 and 6 are also eliminated.
We are left with digits 4 and 8. To be divisible by 8, the last 3 digits should be divisible by 8. 444 is not divisible by 8, so we are now left with an n digit number containing 8's and we need to ensure that the number formed is divisible by 7 and 9.
So we can have 9 8's or 18 8's
Both these numbers are divisible by 9 (sum of the digits should be divisible by 9)
To divisibility test for 7 is the number of 10's + 5 times the units place should be divisible by 7 (the number of 10's is the number up to the 10's place, for e.g. in the number 543, the number of 10's is 54. In the number 6425 it is 642)
8 8 8 8 8 8 8 8 8 = 88888888 + (5 * 8) = 88888928
Now this can be repeated and made smaller by using the same method as above = 8888892 + 8*5 = 8888932
= 888893 + (2 * 5) = 888903
= 88890 + (5 * 3) = 88905
= 8890 + (5 * 3) = 8905
= 890 + (5 * 3) = 905
= 90 + (5 * 3) = 105
= 10 + (5 * 3) = 25 and we know that 25 is not divisible by 7.
Now it is very obvious that the method is cumbersome and time consuming just to find out if the number is divisible by 7. Therefore the better method, would be to directly divide by 7.
Option E
Arun Kumar
