If a, b, and c are integers such that 0

Question

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

Bunuel · Answer

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.

Zarrolou · Answer

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
eq{4}   2b
eq{9}   2b
eq{14}   2b
eq{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
eq{14}   b
eq{7}  ( all other value of b are decimals of out of range 5-10)
 So a=5  b
eq{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.

skamal7 · Answer

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?

Bunuel · Answer

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.

Hope it helps.

khar · Answer

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.

Bunuel · Answer

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

Hope it helps.

stne · Answer

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?

Bunuel · Answer

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.

stne · Answer

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"

Bunuel · Answer

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

aditiphadnis · Answer

How did you arrive at 7 and 8?

As per the table below, the number will not be div if b=7 and c=8 or b= 8 and c=9 and b=6 and c= 9 
 https://gmatclub.com/forum/download/file.php?mode=view&id=54915

Also please explain how did you arrive at 875 here?

Bunuel · Answer

I think you misunderstood the question. The question asks whether a*b*c is divisible by 3, not a + b + c.

sarajane · Answer

Hi,

Can I understand why the answer is not C?

My rationale is that Statement (1) alone is not conclusive as
(i) a = 5 is undisputed.
(ii) b = 4 or 3
(iii) c = 3 or 2

In the first case where a = 5, b = 4 and c = 3, the fraction will be  \frac{1}{200}  +  \frac{4}{100}  +  \frac{3}{10}  =  \frac{69}{200}  (Simplest form).
In the second case where a = 5, b = 4 and c = 2, the fraction  \frac{1}{200}  +  \frac{4}{100}  +  \frac{2}{10}  =  \frac{49}{200}  (Simplest form). 
In the third case where a = 5, b = 3 and c = 2, the fraction  \frac{1}{200}  +  \frac{3}{100}  +  \frac{2}{10}  =  \frac{47}{200}  (Simplest form).

Given that there are 2 possible scenarios: abc is divisible by 3 (Case 1 & 3) and abc is not divisible by 3 (Case 2), only by looking at Statement (2) will we be able to figure out what the value of b and c is and thereby figure out if abc is divisible by 3.

Am I missing something here?

-sj

chetan2u · Answer

For product a*b*c to be divisible by 3, at least one of the three integers should be 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}=\frac{x}{200} 
100c+10b+a=c00+b0+a=cba
So cba and 1000 have only factor 5 in common. 
 This tells us that 100c+10b+a or 3-digit number cba is a multiple of 5 but not of 25 or 2. 
a) 100c+10b+a will be multiple of 5,  only when a is 5  as 0<a<10.
b) Numbers ending in 25, 75 are multiples of 25, so  b cannot be 7 , that is number cba cannot be c75. 
So b can be 6, and c any of the remaining digits 7,8 or 9.  As b is a multiple of 3, product a*b*c will be divisible by 3.  
b can also be 8, then c is 9.  As c is a multiple of 3, product a*b*c will be divisible by 3.

Product a*b*c is always divisible by 3.
Sufficient

(2) c – b < b – a
If c is 8 and b is 7
a) a is 4……answer is No
b) a is 3…..answer is yes
Insufficient

A

bumpbot · Answer

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If a, b, and c are integers such that 0 < a < b < c < 10, is

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