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:

Re: If a, b, and c are integers such that 0 < a < b < c < 10, is [#permalink]
15 Apr 2013, 02:52

7

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

If a, b, and c are integers such that 0 < a < b < c < 10, is the product abc divisible by 3?

(1) If \(\frac{a}{1000}\) + \(\frac{b}{100}\) + \(\frac{c}{10}\) is expressed as a single fraction reduced to lowest terms, the denominator is 200.

\(\frac{a}{1000}\) + \(\frac{b}{100}\) + \(\frac{c}{10}=\frac{a+10b+100c}{1000}\). Since when reduced to lowest terms, the denominator is 1000/5=200, then a+10b+100c must be divisible by 5, which implies that a must be divisible by 5. Now, since 0<a<10, then a=5.

Next, abc won't be divisible by 3, if and only, b and c are 7 and 8 respectively (in all other cases b or c will be divisible by 3 since 5<b<c<10), but in this case a+10b+100c=875=25*35 and in this case \(\frac{a+10b+100c}{1000}=\frac{875}{1000}=\frac{7}{8}\), so reduced to lowest terms the denominator is 8 not 200 as stated.

Therefore abc IS divisible by 3. Sufficient.

(2) c – b < b – a. This implies that a+c<2b. If a=1, b=4 and c=5, then the answer is NO but if a=1, b=6 and c=7, then the answer is YES. Not sufficient.

Re: If a, b, and c are integers such that 0 < a < b < c < 10, is [#permalink]
15 Apr 2013, 03:02

2

This post received KUDOS

emmak wrote:

If a, b, and c are integers such that 0 < a < b < c < 10, is the product abc divisible by 3?

(1) If \(\frac{a}{1000}\) + \(\frac{b}{100}\) + \(\frac{c}{10}\) is expressed as a single fraction reduced to lowest terms, the denominator is 200.

(2) c – b < b – a

A++ to the question! is the product abc divisible by 3? means is at least one a multiple of 3?

\(\frac{a}{1000}\) + \(\frac{b}{100}\) + \(\frac{c}{10}\) expressed as one fraction is \(\frac{a+10b+100c}{1000}\) the factors of 200 are 2*5*2*5*2. To get the fraction to a 200 Den the sum must be a multiple of 5 and must NOT have a 2 or more 5s as factor, otherwise other semplification will be possbile.

1)the sum must be a multiple of 5, \(a+10b+100c\) if this is a multiple of 5 must end in 0 or 5. (note that b will be the first digit of the tens and c will be the first digit of the hundreds and c is the unit). To end in 0 or 5 a must be 0 or 5. a cannot be 0 (0<a) so \(a=5\). Good

2) \(5+10b+100c\) must NOT have a 2 as factor or any more 5. divide by 5 \(1+2b+20c\) what remains after the first division MUST not be even or a multiple of 5. This means that \(2b\neq{4}\) \(2b\neq{9}\) \(2b\neq{14}\) \(2b\neq{19}\) and so on otherwise it will be divisibe: ie 2b=9 1+9+2C will be divisibe by 2 and 5. Of all the values b cannot assume there is one that is interesting : \(2b\neq{14}\) \(b\neq{7}\) ( all other value of b are decimals of out of range 5-10) So a=5 \(b\neq{7}\). With this info every combination abc will have a multiple of 3. The statement is SUFFICIENT.

(2) c – b < b – a \(c+a<2b\) Not sufficient. b=3 c=5 a =1 YES b=4 c=5 a=1 NO. _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: If a, b, and c are integers such that 0 < a < b < c < 10, is [#permalink]
17 Apr 2013, 05:10

Hi bunnel,

Can you plesae explain the below part in ur post little more but in this case a+10b+100c=875=25*35 and in this case \frac{a+10b+100c}{1000}=\frac{875}{1000}=\frac{7}{8}, so reduced to lowest terms the denominator is 8 not 200 as stated.

I a not able to understand how 25*35 comes and also how from the fraction 7/8 ur deducing that abc is divisble by 3? _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Re: If a, b, and c are integers such that 0 < a < b < c < 10, is [#permalink]
17 Apr 2013, 05:28

2

This post received KUDOS

Expert's post

skamal7 wrote:

Hi bunnel,

Can you plesae explain the below part in ur post little more but in this case a+10b+100c=875=25*35 and in this case \frac{a+10b+100c}{1000}=\frac{875}{1000}=\frac{7}{8}, so reduced to lowest terms the denominator is 8 not 200 as stated.

I a not able to understand how 25*35 comes and also how from the fraction 7/8 ur deducing that abc is divisble by 3?

We have that a=5. We also know that 5 < b < c < 10. Now, abc won't be divisible by 3, if and only, b and c are 7 and 8 respectively (in all other cases either b is 6 or c is 9 since 5<b<c<10). So, if we can prove that b and c are NOT 7 and 8 respectively, then abc WILL be divisible by 3.

If b=7 and c=8, then a+10b+100c=875 (875=25*35) and in this case \(\frac{a+10b+100c}{1000}=\frac{875}{1000}=\frac{7}{8}\), so reduced to lowest terms the denominator is 8 not 200 as stated.

Thus, b and c are NOT 7 and 8 respectively. Therefore b is 6 or/and c is 9, so abc IS divisible by 3.

Re: If a, b, and c are integers such that 0 < a < b < c < 10, is [#permalink]
23 Apr 2013, 11:11

Bunuel wrote:

If a, b, and c are integers such that 0 < a < b < c < 10, is the product abc divisible by 3?

(1) If \(\frac{a}{1000}\) + \(\frac{b}{100}\) + \(\frac{c}{10}\) is expressed as a single fraction reduced to lowest terms, the denominator is 200.

\(\frac{a}{1000}\) + \(\frac{b}{100}\) + \(\frac{c}{10}=\frac{a+10b+100c}{1000}\). Since when reduced to lowest terms, the denominator is 1000/5=200, then a+10b+100c must be divisible by 5, which implies that a must be divisible by 5. Now, since 0<a<10, then a=5.

Next, abc won't be divisible by 3, if and only, b and c are 7 and 8 respectively (in all other cases b or c will be divisible by 3 since 5<b<c<10), but in this case a+10b+100c=875=25*35 and in this case \(\frac{a+10b+100c}{1000}=\frac{875}{1000}=\frac{7}{8}\), so reduced to lowest terms the denominator is 8 not 200 as stated.

Therefore abc IS divisible by 3. Sufficient.

(2) c – b < b – a. This implies that a+c<2b. If a=1, b=4 and c=5, then the answer is NO but if a=1, b=6 and c=7, then the answer is YES. Not sufficient.

Answer: A.

Hope it's clear.

HI,

could you please explain how (100C+10b+a)/1000 has a denominator with 200? as when u take a three digit number we express it as 100C+10B+A, So how a three digit number when divided by 1000 has 200 as denominator? i thought E as the answer (please correct me if i am wrong) .

Re: If a, b, and c are integers such that 0 < a < b < c < 10, is [#permalink]
24 Apr 2013, 04:32

1

This post received KUDOS

Expert's post

khar wrote:

Bunuel wrote:

If a, b, and c are integers such that 0 < a < b < c < 10, is the product abc divisible by 3?

(1) If \(\frac{a}{1000}\) + \(\frac{b}{100}\) + \(\frac{c}{10}\) is expressed as a single fraction reduced to lowest terms, the denominator is 200.

\(\frac{a}{1000}\) + \(\frac{b}{100}\) + \(\frac{c}{10}=\frac{a+10b+100c}{1000}\). Since when reduced to lowest terms, the denominator is 1000/5=200, then a+10b+100c must be divisible by 5, which implies that a must be divisible by 5. Now, since 0<a<10, then a=5.

Next, abc won't be divisible by 3, if and only, b and c are 7 and 8 respectively (in all other cases b or c will be divisible by 3 since 5<b<c<10), but in this case a+10b+100c=875=25*35 and in this case \(\frac{a+10b+100c}{1000}=\frac{875}{1000}=\frac{7}{8}\), so reduced to lowest terms the denominator is 8 not 200 as stated.

Therefore abc IS divisible by 3. Sufficient.

(2) c – b < b – a. This implies that a+c<2b. If a=1, b=4 and c=5, then the answer is NO but if a=1, b=6 and c=7, then the answer is YES. Not sufficient.

Answer: A.

Hope it's clear.

HI,

could you please explain how (100C+10b+a)/1000 has a denominator with 200? as when u take a three digit number we express it as 100C+10B+A, So how a three digit number when divided by 1000 has 200 as denominator? i thought E as the answer (please correct me if i am wrong) .

Khar.

For example, if a=5, b=6 and c=7, then \(\frac{a+10b+100c}{1000}=\frac{765}{1000}=\frac{153}{200}\).

Re: If a, b, and c are integers such that 0 < a < b < c < 10, is [#permalink]
21 Jul 2013, 22:36

This is a good question, but the part where we have a+10b+100c and which implies that a is the unit digit is not clear to me . What is the concept here? How are we able to deduce that a is the unit digit , b tens and c hundreds? _________________

Re: If a, b, and c are integers such that 0 < a < b < c < 10, is [#permalink]
21 Jul 2013, 22:47

Expert's post

stne wrote:

This is a good question, but the part where we have a+10b+100c and which implies that a is the unit digit is not clear to me . What is the concept here? How are we able to deduce that a is the unit digit , b tens and c hundreds?

Any 3-digit number XYZ can be represented as 100X + 10Y + Z, for example 246 = 2*100 + 4*10 + 6.

Since, a, b, and c are single digits (0 < a < b < c < 10), then 100c + 10b + a gives a 3-digit integer cba (the same way as above).

Re: If a, b, and c are integers such that 0 < a < b < c < 10, is [#permalink]
21 Jul 2013, 23:06

Bunuel wrote:

stne wrote:

This is a good question, but the part where we have a+10b+100c and which implies that a is the unit digit is not clear to me . What is the concept here? How are we able to deduce that a is the unit digit , b tens and c hundreds?

Any 3-digit number XYZ can be represented as 100X + 10Y + Z, for example 246 = 2*100 + 4*10 + 6.

Since, a, b, and c are single digits (0 < a < b < c < 10), then 100c + 10b + a gives a 3-digit integer cba (the same way as above).

Hope it's clear.

yups now its clear

Zarrolou's comment as highlighted below confused me,I guess he meant b will be the tens digit and c will be the hundreds digit and a will be the units digit, "b will be the first digit of the tens" did not make sense to me. Is that possible? b will be the tens digit, what do we mean by " first digit of the tens and first digit of the Hundreds"

Zarrolou wrote:

".... (note that b will be the first digit of the tens and c will be the first digit of the hundreds and c is the unit)...."

Re: If a, b, and c are integers such that 0 < a < b < c < 10, is [#permalink]
21 Jul 2013, 23:09

1

This post received KUDOS

Expert's post

stne wrote:

Bunuel wrote:

stne wrote:

This is a good question, but the part where we have a+10b+100c and which implies that a is the unit digit is not clear to me . What is the concept here? How are we able to deduce that a is the unit digit , b tens and c hundreds?

Any 3-digit number XYZ can be represented as 100X + 10Y + Z, for example 246 = 2*100 + 4*10 + 6.

Since, a, b, and c are single digits (0 < a < b < c < 10), then 100c + 10b + a gives a 3-digit integer cba (the same way as above).

Hope it's clear.

yups now its clear

Zarrolou's comment as highlighted below confused me,I guess he meant b will be the tens digit and c will be the hundreds digit and a will be the units digit, "b will be the first digit of the tens" did not make sense to me. Is that possible? b will be the tens digit, what do we mean by " first digit of the tens and first digit of the Hundreds"

Zarrolou wrote:

".... (note that b will be the first digit of the tens and c will be the first digit of the hundreds and c is the unit)...."

I think he meant a=units, b=tens, and c=hundreds. _________________

Re: If a, b, and c are integers such that 0 < a < b < c < 10, is [#permalink]
09 Aug 2014, 10:53

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. _________________

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...