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

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15 Apr 2013, 03:52

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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]

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17 Apr 2013, 06:28

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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]

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15 Apr 2013, 04:02

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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]

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24 Apr 2013, 05:32

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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]

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22 Jul 2013, 00:09

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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]

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17 Apr 2013, 06:10

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]

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23 Apr 2013, 12:11

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]

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21 Jul 2013, 23:36

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]

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21 Jul 2013, 23:47

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]

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22 Jul 2013, 00:06

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]

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09 Aug 2014, 11:53

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

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