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Is the 3-digit number abc divisible by 7? ( a , b , and c [#permalink]
01 Jan 2004, 21:06
00:00
A
B
C
D
E
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
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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. _________________
Best,
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
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? _________________
Best,
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
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.
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.
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.
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? _________________
Best,
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
(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.... _________________
Best,
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
(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. _________________
Best,
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