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Divisibility

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Divisibility

by chieftang » Tue Jan 03, 2012 12:05 pm
Given a 5-digit number abcde, if 2d + e = 16 then which of the following must be true:

I. The number abcde is divisible by 2.
II. The number abcde is divisible by 4.
III. The number abcde is divisible by 8.

(A) I only
(B) I, II only
(C) II, III only
(D) I, II, III
(E) None must be true


Source: MOEMS problem bank
Last edited by chieftang on Tue Jan 03, 2012 2:33 pm, edited 1 time in total.
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by shankar.ashwin » Tue Jan 03, 2012 12:33 pm
Given 2d + e = 16

2d is even, hence e is even too and e can take values 0,2,4,6 and 8.

'd' can take values 4,5,6,7 and 8.

when d = 4, e = 8
when d = 5, e = 6
when d = 6, e = 4
when d = 7, e = 2
when d = 8, e = 0

Divisibility rule for 2 - 'e' should be even - All options satisfy
Divisibility rule for 4 - 'de' should be divisible by 4 - All options satisfy
Divisibility rule for 8 - 'cde' should be divisible by 8 - We dont know about 'c' - Cant say

[spoiler](B) I, II only [/spoiler]
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by chufus » Tue Jan 03, 2012 2:14 pm
I think the answer should be A..

Even with reference to your explanation:

when d = 5, e = 6

so de = 30 which is not divisible by 4........ All the question tells us is that e is even by saying 2d + e = 16

So e is always an even integer >= 0 And that simply means that abcde is even. Now there are a host of even numbers that are divisible by 2 always, but not divisible by 4 or 8.

Somehow I feel the answer should be A

Just checked for divisibility rules and yes you are right.... I retract my explanation....

However, this question actually relies on knowing beforehand the divisibility rule for 4. Does GMAT really expect people to do so?
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by neelgandham » Tue Jan 03, 2012 2:28 pm
Chufus, I do agree to the point that the question tells us e is even by saying 2d + e = 16. However it also indicates that abcde can be one among

abc80
abc72
abc64
abc56
abc48

Because
when d = 4, e = 8
when d = 5, e = 6
when d = 6, e = 4
when d = 7, e = 2
when d = 8, e = 0

abc80 = abc00 + 80 - Is divisible by 2
abc72 = abc00 + 72 - Is divisible by 2
abc64 = abc00 + 64 - Is divisible by 2
abc56 = abc00 + 56 - Is divisible by 2
abc48 = abc00 + 48 - Is divisible by 2

abc80 = abc00 + 80 - Is divisible by 4 because abc00 is divisible by 4 and 80 is also divisible by 4
abc72 = abc00 + 72 - Is divisible by 4 because abc00 is divisible by 4 and 72 is also divisible by 4
abc64 = abc00 + 64 - Is divisible by 4 because abc00 is divisible by 4 and 64 is also divisible by 4
abc56 = abc00 + 56 - Is divisible by 4 because abc00 is divisible by 4 and 56 is also divisible by 4
abc48 = abc00 + 48 - Is divisible by 4 because abc00 is divisible by 4 and 48 is also divisible by 4

abc00 is divisible by 4 because abc00 = abc * 25 * 4, a multiple of 4.

However, all the numbers are divisible by 8 only if abc(three digit number) is an even number or in other words all the numbers are divisible by 8 only if c(unit's digit) is an even number. Since we don't know the value of c, we are not sure if it is divisible by 8

IMO B
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by chufus » Tue Jan 03, 2012 2:37 pm
Good One Mate... Agreed one should be able to decipher that any multiple of 100 is always divisible by 4 .... and the fact that 100 is not divisible by 8 simply because 100 = 2*2*5*5 , for it to be divisible by 8 the last two digits give us 2 factors of 2 but to be sure about the third factor of 2 we need to know c, which we don't, so "MUST" condition is not satisfied.....

Good One ! Rock on ! Just made me more aware of number properties..
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by neelgandham » Tue Jan 03, 2012 2:39 pm
I am glad you found it helpful !:-)
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by chieftang » Tue Jan 03, 2012 2:48 pm
Indeed a good way to look at it, neelgandham!

Here are some mental math tricks you guys might also want to add to the arsenal:

The number abcde is a multiple of 2 if e is a multiple of 2
The number abcde is a multiple of 4 if 2d+e is a multiple of 4
The number abcde is a multiple of 8 if 4c+2d+e is a multiple of 8

Example: 75392 is divisible by 8 because 12+18+2=32, which is divisible by 8.

Answer: B
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by ariz » Wed Jan 04, 2012 9:31 am
You all are quite simply amazing...
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