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Is the 3digit number abc divisible by 7? ( a , b , and c [#permalink]
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01 Jan 2004, 22:06
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This topic is locked. If you want to discuss this question please repost it in the respective forum. Q: Is the 3digit number abc divisible by 7? ( a, b, and c are the respective digits of the 3digit number) (1): The 2digit number ab minus ( c * 2) is divisible by 7. (2): 2* a + 3* b + c is divisible by 7.
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Either is sufficient.
And I cannot explain why...



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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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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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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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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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1. take a =b=3 and c = 1 =>
92=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]
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02 Jan 2004, 12:58
bat_car wrote: 1. take a =b=3 and c = 1 =>
92=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 abc*2 = 332 = 31. It is not a*bc*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]
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02 Jan 2004, 19:49
Geethu wrote: bat_car wrote: 1. take a =b=3 and c = 1 =>
92=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 abc*2 = 332 = 31. It is not a*bc*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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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



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Here is the general solution:
10*a+b2c = 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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Excellent approach mantha. Two thumbs up!
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condition 1)
ab2c = 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 = 350j140b49c so abc is divisible



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Quote: (1): The 2digit 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+b2c)/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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dj wrote: Quote: (1): The 2digit 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+b2c)/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 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



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AkamaiBrah wrote: dj wrote: Quote: (1): The 2digit 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+b2c)/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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dj wrote: AkamaiBrah wrote: dj wrote: Quote: (1): The 2digit 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+b2c)/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










