Given that R is positive three-digit integer, what is the : Data Sufficiency (DS)

mydreammba

Joined: 28 Jul 2011

Last visit: 06 Dec 2013

Posts: 224

Own Kudos:

1,581

[1]

Given Kudos: 16

Location: United States

Concentration: International Business, General Management

GPA: 3.86

WE:Accounting (Commercial Banking)

Posts: 224

Kudos: 1,581

[1]

22 Nov 2011, 06:51

1

Kudos

Bookmarks

Initially made silly mistake

+1 C

thisiszico2006

Joined: 16 Dec 2009

Last visit: 13 Aug 2018

Posts: 55

Own Kudos:

78

Given Kudos: 11

GMAT 1: 680 Q49 V33 Verified score

WE:Information Technology (Commercial Banking)

GMAT 1: 680 Q49 V33 Verified score

Posts: 55

Kudos: 78

25 Nov 2011, 05:53

Kudos

Bookmarks

Made the same silly mistake on both the options..

. Picked (D)

prateekbhatt

Joined: 28 Jun 2011

Last visit: 26 Jul 2019

Posts: 768

Own Kudos:

279

Given Kudos: 57

Status:My Thread Master Bschool Threads-->Krannert(Purdue),WP Carey(Arizona),Foster(Uwashngton)

Products:

Posts: 768

Kudos: 279

25 Nov 2011, 12:03

Kudos

Bookmarks

I too picked D thinking r=870/3......

C +1

siddharthmuzumdar

Joined: 12 Oct 2011

Last visit: 15 Dec 2015

Posts: 110

Own Kudos:

181

Given Kudos: 110

Posts: 110

Kudos: 181

04 Jan 2012, 12:24

Kudos

Bookmarks

Oh nice question. Very tricky. C is the answer. The first statement could lead to multiple values for the hundreds digit. The second statement could also lead to two values for the hundreds digit.
However, together, the two statements can lead only to one value - 9 - for the hundreds digit of the 3-digit number.

shinbhu

Joined: 29 Jul 2011

Last visit: 14 Oct 2013

Posts: 51

Own Kudos:

171

Given Kudos: 6

Location: United States

Posts: 51

Kudos: 171

04 Jan 2012, 14:03

Kudos

Bookmarks

Oops! Got S1 analysis incorrect, S2 correct. I picked A.

For S1, I totally ignored that the number could be 4 digit.

It is these simple, yet tricky, twists that I fear of.

jlgdr

Joined: 06 Sep 2013

Last visit: 24 Jul 2015

Posts: 1,316

Own Kudos:

2,717

[1]

Given Kudos: 355

Concentration: Finance

Schools: Wharton '17 (A) HBS '17 (WA$)

Posts: 1,316

Kudos: 2,717

[1]

27 Mar 2014, 06:16

1

Kudos

Bookmarks

Again let’s think about this for a second. First we have that ABC * 3 has a hundreds digit of 8. This ain't gonna tell us much so insuff. Now,

Second statement says that abc + 1 has hundreds digit of 9. Now the max value for tens and units digit respectively are 9 and 9. Then we only have 1 carried over to the units digit. Therefore a is either 8 or 9.

Now, let's try with both statements together. If we have 8XX * 3 = 4 as hundreds digit and we would need a carry over of 4 to reach 8 in the hundreds digit as per statement 1. This is not possible because even if b = 9 the maximum carry over we can get is 2 (29 if both units digit and tens digit are equal to 9). Therefore this option is not possible. What if hundreds digit is 9? Then we wil have that 3*9=27 so 7 for hundreds digit we would need only a carry over of 1 which is indeed possible. Hence only 9 works. C

Hope this helps
Cheers
J

arindamsur

Joined: 07 Aug 2011

Last visit: 23 Mar 2016

Posts: 29

Own Kudos:

144

[1]

Given Kudos: 10

Status:Learning

Location: India

Schools: WBUT - Class of 2011

GMAT Date: 01-06-2014

GPA: 2.6

WE:Research (Education)

Schools: WBUT - Class of 2011

Posts: 29

Kudos: 144

[1]

03 Oct 2015, 03:58

1

Kudos

Bookmarks

But the question is saying that R is positive three-digit integer

bhaskar438

Joined: 01 Jan 2015

Last visit: 30 Apr 2023

Posts: 54

Own Kudos:

535

Given Kudos: 14

Posts: 54

Kudos: 535

Updated on: 04 Oct 2015, 10:51

Originally posted by bhaskar438 on 03 Oct 2015, 14:10.
Last edited by bhaskar438 on 04 Oct 2015, 10:51, edited 1 time in total.

Kudos

Bookmarks

kannn

Given that R is positive three-digit integer, what is the hundreds digit of R?

1. The hundreds digit of 3R is 8
2. (R+1) results in a number with the hundreds digit of 9.

Source : 800score.com

~~I believe the correct answer to this question should be changed to answer choice A.~~The question stem says R is a 3 digit number.

Let'say R is represented by digits A,B,and C, where A is a digit between 1 and 9 inclusive and B and C are digits between 0 and 9 inclusive.

(1) R=ABC can be written as R=100*A+10*B+C. 3*R can be written as 300*A+30*B+3*C. The maximum value of 30*B+3*C is 297 when B and C are each 9.
Since it is required that 300*A+30*B+3*C > 800, if we minimize A we get, 300*A > 503. But A cannot be greater than or equal to 3, because then the number would be at least 900 and the hundreds digit of 3R will no longer be 8. So A must be 2. First statement is sufficient

(2) You can easily find this statement insufficient by taking numbers 899 and 900. The hundreds digit of R can be 8 or 9, hence insufficient.

~~Correct answer choice is A.~~

bhaskar438

Joined: 01 Jan 2015

Last visit: 30 Apr 2023

Posts: 54

Own Kudos:

535

Given Kudos: 14

Posts: 54

Kudos: 535

03 Oct 2015, 14:15

Kudos

Bookmarks

VeritasPrepKarishma

kannn

Given that R is positive three-digit integer, what is the hundreds digit of R?

1. The hundreds digit of 3R is 8
2. (R+1) results in a number with the hundreds digit of 9.

Source : 800score.com

Responding to a pm:

1. The hundreds digit of 3R is 8

When you multiply a number by 3, the hundreds digit will be obtained by multiplying the hundreds digit of the original number by 3 and adding any carryover we might have. 3 * 9 (the largest digit) is 27 so the maximum carryover we can get from tens digit multiplication is 2 (even if the units digit of the original number is also 9, we will get 2 as carryover from units to tens and still 2 as carryover from tens to hundreds).

No carry over: 6 times 3 is 18.
611 ---> 611 * 3 = 1833

One carry over: We need to obtain 7 in the hundreds place. 9 times 3 is 27.
961 ---> 961 * 3 = 2883

Two carry over: We need to obtain 6 in hundreds place. 2 times 3 is 6.
291 ---> 291 * 3 = 873

The hundreds digit can be 2, 6 or 9.

2. (R+1) results in a number with the hundreds digit of 9.

R can be an number from 899 to 998.

The hundreds digit can be 8 or 9.

Using both statements, we see that the only common hundreds digit we have is 9. Hence the hundreds digit must be 9.

Answer (C)

Hi VeritasPrepKarishma,

It is given that R is a 3 digit positive integer. Shouldn't the correct answer be choice A?

KarishmaB

Tutor

Joined: 16 Oct 2010

Last visit: 10 Jul 2025

Posts: 16,101

Own Kudos:

74,242

[1]

Given Kudos: 475

Location: Pune, India

Expert

Expert reply

Posts: 16,101

Kudos: 74,242

[1]

04 Oct 2015, 00:41

1

Kudos

Bookmarks

bhaskar438

VeritasPrepKarishma

kannn

Given that R is positive three-digit integer, what is the hundreds digit of R?

1. The hundreds digit of 3R is 8
2. (R+1) results in a number with the hundreds digit of 9.

Source : 800score.com

Responding to a pm:

1. The hundreds digit of 3R is 8

When you multiply a number by 3, the hundreds digit will be obtained by multiplying the hundreds digit of the original number by 3 and adding any carryover we might have. 3 * 9 (the largest digit) is 27 so the maximum carryover we can get from tens digit multiplication is 2 (even if the units digit of the original number is also 9, we will get 2 as carryover from units to tens and still 2 as carryover from tens to hundreds).

No carry over: 6 times 3 is 18.
611 ---> 611 * 3 = 1833

One carry over: We need to obtain 7 in the hundreds place. 9 times 3 is 27.
961 ---> 961 * 3 = 2883

Two carry over: We need to obtain 6 in hundreds place. 2 times 3 is 6.
291 ---> 291 * 3 = 873

The hundreds digit can be 2, 6 or 9.

2. (R+1) results in a number with the hundreds digit of 9.

R can be an number from 899 to 998.

The hundreds digit can be 8 or 9.

Using both statements, we see that the only common hundreds digit we have is 9. Hence the hundreds digit must be 9.

Answer (C)

Hi VeritasPrepKarishma,

It is given that R is a 3 digit positive integer. Shouldn't the correct answer be choice A?

R is a three digit number but 3R needn't be. It could have 4 digits.
In each of the cases shown, R has 3 digits and the hundreds digit is different. So stmnt 1 is not enough.

bhaskar438

Joined: 01 Jan 2015

Last visit: 30 Apr 2023

Posts: 54

Own Kudos:

535

Given Kudos: 14

Posts: 54

Kudos: 535

04 Oct 2015, 11:21

Kudos

Bookmarks

VeritasPrepKarishma

R is a three digit number but 3R needn't be. It could have 4 digits.
In each of the cases shown, R has 3 digits and the hundreds digit is different. So stmnt 1 is not enough.

Yes you are right! Perhaps it is best to avoid an algebraic approach such as the one I did for a multiplication problem such as this. I will leave that algebraic approach for addition and subtraction questions.

+1 Kudos to you.

bhaskar438

Joined: 01 Jan 2015

Last visit: 30 Apr 2023

Posts: 54

Own Kudos:

535

Given Kudos: 14

Posts: 54

Kudos: 535

04 Oct 2015, 11:24

Kudos

Bookmarks

kannn

Given that R is positive three-digit integer, what is the hundreds digit of R?

1. The hundreds digit of 3R is 8
2. (R+1) results in a number with the hundreds digit of 9.

Source : 800score.com

Here is an alternate method to show statement 1 is insufficient without looking at carryover. 3R is a multiple of 3, therefore its sum of digits should be a multiple of 3. In the case 3R is a 4 digit number, the thousands digit can only be 1 or 2 (the largest 3 digit number, 999, multiplied by 3 is smaller than 3000). After applying statement 1, the hundreds digit is now 8.

Let's say 3R is represented by XYZW, where X has a digit value of 1 or 2, Y has a digit value of 8, and Z and W each can be a digit between 0 and 9 inclusive.

Case 1: X=1 and Y=8

1+8+Z+W = Multiple of 3
Ex: If Z+W = 0. Then R = 600. The hundreds digit of R is 6.

Case 2: X=2 and Y=8

2+8+Z+W = Multiple of 3
Ex: If Z and W each equal 1, then 3R =2811 --> R=927. The hundreds digit of R is 9.

Statement 1 is insufficient.

KarishmaB

Tutor

Joined: 16 Oct 2010

Last visit: 10 Jul 2025

Posts: 16,101

Own Kudos:

74,242

Given Kudos: 475

Location: Pune, India

Expert

Expert reply

Posts: 16,101

Kudos: 74,242

01 Jan 2019, 07:08

Kudos

Bookmarks

VeritasKarishma

kannn

Given that R is positive three-digit integer, what is the hundreds digit of R?

1. The hundreds digit of 3R is 8
2. (R+1) results in a number with the hundreds digit of 9.

Source : 800score.com

Responding to a pm:

1. The hundreds digit of 3R is 8

When you multiply a number by 3, the hundreds digit will be obtained by multiplying the hundreds digit of the original number by 3 and adding any carryover we might have. 3 * 9 (the largest digit) is 27 so the maximum carryover we can get from tens digit multiplication is 2 (even if the units digit of the original number is also 9, we will get 2 as carryover from units to tens and still 2 as carryover from tens to hundreds).

No carry over: 6 times 3 is 18.
611 ---> 611 * 3 = 1833

One carry over: We need to obtain 7 in the hundreds place. 9 times 3 is 27.
961 ---> 961 * 3 = 2883

Two carry over: We need to obtain 6 in hundreds place. 2 times 3 is 6.
291 ---> 291 * 3 = 873

The hundreds digit can be 2, 6 or 9.

2. (R+1) results in a number with the hundreds digit of 9.

R can be an number from 899 to 998.

The hundreds digit can be 8 or 9.

Using both statements, we see that the only common hundreds digit we have is 9. Hence the hundreds digit must be 9.

Answer (C)

Quote:

For this question, what do you mean 1 carry over and 2 carry over, and how you get those numbers?

Say R = XYZ

What is 3R?

3R = 3 * XYZ

Say XYZ = 456

How will you multiply it by 3?

3 * 456 =>
3 times 6 is 18 so we write 8 in units digit and 1 carry over. We get ......8
3 times 5 is 15. Add 1 from carry over to get 16. Now write 6 in tens digit and 1 carry over. We get .......68
3 times 4 is 12. Add 1 from carry over to get 13. Now write 13. We get 1368

So how did we get the 3 in the hundreds place of 1368? We multiplied 3 with the hundreds digit (X = 4) to get 12. To that we added 1 (from carry over) and we got 13. So 3 came in our hundreds place.

This is exactly what we are doing. From previous calculations, the carry over can be 0/1/2 (explained above).
So we take each case. If carry over were 0, then R could have been something like 611 or 622 or 613 etc. Why? Because to get 8 in hundreds place, 3 needs to be multiples by 6. So 6 should be X. Y and Z should be numbers which give no carry over complication so 1/2/3 work.

We did the same thing for 1 carry over and 2 carry over too.

Sri07

Joined: 27 Jun 2019

Last visit: 30 Aug 2021

Posts: 47

Own Kudos:

9

Given Kudos: 167

Posts: 47

Kudos: 9

29 Sep 2020, 05:15

Kudos

Bookmarks

I am surely gonna skip this type of question on the exam!

bumpbot

Non-Human User

Joined: 09 Sep 2013

Last visit: 04 Jan 2021

Posts: 37,375

Own Kudos:

1,010

Posts: 37,375

Kudos: 1,010

10 Dec 2022, 07:21

Kudos

Bookmarks

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

Given that R is positive three-digit integer, what is the

Prep Toolkit

My Rewards

GMAT Data Sufficiency (DS) Questions

Given that R is positive three-digit integer, what is the

Prep Toolkit

My Rewards