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Is the 3-digit number abc divisible by 7? ( a , b , and c

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Is the 3-digit number abc divisible by 7? ( a , b , and c [#permalink] New post 01 Jan 2004, 21:06
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A
B
C
D
E

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Q: Is the 3-digit number abc divisible by 7? (a, b, and c are the respective digits of the 3-digit number)

(1): The 2-digit number ab minus (c * 2) is divisible by 7.
(2): 2*a + 3*b + c is divisible by 7.
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 [#permalink] New post 01 Jan 2004, 21:56
Either is sufficient.

And I cannot explain why...
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 [#permalink] New post 01 Jan 2004, 22:10
stoolfi wrote:
Either is sufficient.

And I cannot explain why...


Initially I didn't find

(2): 2*a + 3*b + c is divisible by 7 useful

For example,
112 = 2(1)+3(1)+2 =7
497 = 2(4)+3(9)+7 =42

While I tried with the example here I found this to be sufficient. You are right.

Answer should be D.
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 [#permalink] New post 01 Jan 2004, 22:25
Geethu wrote:
stoolfi wrote:
Either is sufficient.

And I cannot explain why...


Initially I didn't find

(2): 2*a + 3*b + c is divisible by 7 useful

For example,
112 = 2(1)+3(1)+2 =7
497 = 2(4)+3(9)+7 =42

While I tried with the example here I found this to be sufficient. You are right.



Answer should be D.


Are you SURE that this is true for ALL a, b, and c?
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 [#permalink] New post 01 Jan 2004, 22:30
I suppose a student who had worked very hard would recognize both of these tests as divisibility tests for the number 7. In fact, when I looked at them, each looked familiar, since I had gone off on a little tangent about this after missing an only slightly related problem months ago.

To "solve" this problem, I tried to pick a few very different numbers that fit the case-- odds and evens, all odds and all evens, 007, 056, etc. I also wrote out a string of the numbers, and it started to "clixk" mentally as I saw how changing the digits changed the formulas.

Time spent: 3 minutes. Too much for the test. To be really sharp, I suppose, one would have to quickly identify a reason why these formulas might work. I couldn't do it, and I hope someone who knew the question intuitively can explain it.

On the other hand, maybe I missed it...
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 [#permalink] New post 02 Jan 2004, 08:20
Is there a trick in this ? I don't find a 3 digit number which doesn't follow the formula 2.
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Ans E [#permalink] New post 02 Jan 2004, 11:51
1. take a =b=3 and c = 1 =>

9-2=7 but 3.3.1 is not divi by 7
=> 1 insuff

2. take a=3, b=1 and c=5 =>14/7 is good but a.b.c is not % 7.

for C - considering 1 and 2 together does not lead any where either. Rather I just looked at the given equations and my gut feel is that they will not, hence Ans = E.
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Re: Ans E [#permalink] New post 02 Jan 2004, 11:58
bat_car wrote:
1. take a =b=3 and c = 1 =>

9-2=7 but 3.3.1 is not divi by 7
=> 1 insuff

2. take a=3, b=1 and c=5 =>14/7 is good but a.b.c is not % 7.

for C - considering 1 and 2 together does not lead any where either. Rather I just looked at the given equations and my gut feel is that they will not, hence Ans = E.


We will take your example itself.
1. If we take a=b=3 then ab-c*2 = 33-2 = 31. It is not a*b-c*2

2. a=3,b=1,c=5 then abc= 315 which is divisible by 7.

The problem says abc are digits of a three digit number.
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Re: Ans E [#permalink] New post 02 Jan 2004, 18:49
Geethu wrote:
bat_car wrote:
1. take a =b=3 and c = 1 =>

9-2=7 but 3.3.1 is not divi by 7
=> 1 insuff

2. take a=3, b=1 and c=5 =>14/7 is good but a.b.c is not % 7.

for C - considering 1 and 2 together does not lead any where either. Rather I just looked at the given equations and my gut feel is that they will not, hence Ans = E.


We will take your example itself.
1. If we take a=b=3 then ab-c*2 = 33-2 = 31. It is not a*b-c*2

2. a=3,b=1,c=5 then abc= 315 which is divisible by 7.

The problem says abc are digits of a three digit number.


It is easy to find examples that work. Can anyone figure out a simple way to PROVE that either statement is sufficient for ALL a, b, and c?
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yep.. [#permalink] New post 02 Jan 2004, 20:12
Here is the general solution:

10*a+b-2c = 7k (k any positive integer)

100a+10b+c = 10(10a+b) +c = 10(7k+2c)+c = 7{(10k)+3c} is div by 7.

Do the next on the same lines but substitute for c instead and that will give an answer as well.
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 [#permalink] New post 03 Jan 2004, 07:20
Excellent approach mantha. Two thumbs up! :band :band
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 [#permalink] New post 10 Jan 2004, 16:10
condition 1)

ab-2c = 7k
10a+b = 7k+2c
so
100a+10b+c = 70k+20c+c = 7(10k+3c) so abc is divisible

condition 2)

2a+3b+c = 7j

100a+150b+50c = 350j
so
100a+10b+c = 350j-140b-49c so abc is divisible
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 [#permalink] New post 10 Jan 2004, 17:20
Quote:
(1): The 2-digit number ab minus (c * 2) is divisible by 7.
(2): 2*a + 3*b + c is divisible by 7.


got D. this is how:

to prove: abc/7 or, (100a+10b+c)/7

1) (10a+b-2c)/7 = integer
or, 10a/7 + b/7 -2c/7 = integer
each of the value needs to be a multiple of 7 to make the final answer an integer.

which tells that (100a+10b+c)/7 is an integer
so, sufficient

2)2a/7 + 3b/7 + c/7 = integer
same reasoning as 1st's

sufficient.
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 [#permalink] New post 11 Jan 2004, 03:43
dj wrote:
Quote:
(1): The 2-digit number ab minus (c * 2) is divisible by 7.
(2): 2*a + 3*b + c is divisible by 7.


got D. this is how:

to prove: abc/7 or, (100a+10b+c)/7

1) (10a+b-2c)/7 = integer
or, 10a/7 + b/7 -2c/7 = integer
each of the value needs to be a multiple of 7 to make the final answer an integer.

which tells that (100a+10b+c)/7 is an integer
so, sufficient

2)2a/7 + 3b/7 + c/7 = integer
same reasoning as 1st's

sufficient.


This is bad reasoning. A sum can be divisible by something that none of the part are divisible by....
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AkamaiBrah
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Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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 [#permalink] New post 11 Jan 2004, 05:36
AkamaiBrah wrote:
dj wrote:
Quote:
(1): The 2-digit number ab minus (c * 2) is divisible by 7.
(2): 2*a + 3*b + c is divisible by 7.


got D. this is how:

to prove: abc/7 or, (100a+10b+c)/7

1) (10a+b-2c)/7 = integer
or, 10a/7 + b/7 -2c/7 = integer
each of the value needs to be a multiple of 7 to make the final answer an integer.

which tells that (100a+10b+c)/7 is an integer
so, sufficient

2)2a/7 + 3b/7 + c/7 = integer
same reasoning as 1st's

sufficient.


This is bad reasoning. A sum can be divisible by something that none of the part are divisible by....


accepted. But, won't individual divisibilities lead to Sum's divisibility?
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 [#permalink] New post 14 Jan 2004, 00:36
dj wrote:
AkamaiBrah wrote:
dj wrote:
Quote:
(1): The 2-digit number ab minus (c * 2) is divisible by 7.
(2): 2*a + 3*b + c is divisible by 7.


got D. this is how:

to prove: abc/7 or, (100a+10b+c)/7

1) (10a+b-2c)/7 = integer
or, 10a/7 + b/7 -2c/7 = integer
each of the value needs to be a multiple of 7 to make the final answer an integer.

which tells that (100a+10b+c)/7 is an integer
so, sufficient

2)2a/7 + 3b/7 + c/7 = integer
same reasoning as 1st's

sufficient.


This is bad reasoning. A sum can be divisible by something that none of the part are divisible by....


accepted. But, won't individual divisibilities lead to Sum's divisibility?


Yes, but you claim in your reasoning that each part "needs to be" divisible in order for the sum to be divisible. this is not true.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

  [#permalink] 14 Jan 2004, 00:36
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