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bugigio
08-30-2004, 05:59 PM
Hi all,
I downloaded from the ETS website the GRE Powerprep 3.1 version and I've just come across with some questions I can't solve. Would you please help me?

1) The reflection of a positive integer is obtained by reversing its digits. For ex. 321 is the reflection of 123. The difference between a five-digit integer and its reflection must be divisible by...... 2,4,5,6,9.

The correct answer is 9 but I can't realise why. I must say I have also problems to identify a 5 digit integer.... (is there only 12345 o also some others such as all the numbers made of 5 digits?)

2) 7 one $ bills are to be distributed among Lucia, Gomez and DOmingo so that each person receive at least 1$.
A) The number of ways to distribute the bills so that at least one of the persons receive at leat 3$
B) The total number of ways to disribute the bills
CORRECT ANSWER: A and B are equal Why? :hmm:

3) 3<x<5
A) (3+x)/3
B) (5+x)/5
CORRECT ANSWER: Can't be stated...
But....if x=3+eps then A)=2+eps and B)=(8/5)+eps then A)>B)
...if x=5-eps then A)=(8/3)-eps and B)=2-eps then A)>B)
...if x=4 then A)=7/3 and B)=9/5 then A)>B)
So why isn't "A is greater than B" the correct answer? :hmm: :hmm:

bigduke
08-31-2004, 12:52 AM
Ans 1) divisibility rule for 9, remember that ?
Ans 2) make combinations, write them down if need be both columns turn out to be 6.
Ans 3) Ah this is a tricky one, most people like you (me included) don't think beyond the integer realm in the first glance. Consider values that are very close to 3 eg 3.01 and then consider values closer to 5 eg 4.99, observe any anomaly?

bugigio
08-31-2004, 09:30 AM
Ans 1) divisibility rule for 9, remember that ?
Ans 2) make combinations, write them down if need be both columns turn out to be 6.
Ans 3) Ah this is a tricky one, most people like you (me included) don't think beyond the integer realm in the first glance. Consider values that are very close to 3 eg 3.01 and then consider values closer to 5 eg 4.99, observe any anomaly?

Ans 1) Divisibility rule for nine? Can you please explain me both this rule and the solution of the problem?
Ans 2) Combinations and permutations are waht I know less.... I m gonna study them harder...
Ans 3).....let me try......

lanpapa
08-31-2004, 09:55 AM
1. if u use 12345, the answer would be 2,4,6,9 so try to use random number for example 13472 or else, then you'll got the number, 9.

2. bigduke, could you please explain it more detail?
A) ?
B) Combination of 7 to 3?

3. i know most of this type of questions is "D" (hmm, based on my experience)
but this one, !!! geez, after i look the problem several times, and try to figure bigduke's answer, i found myself looking the problem in powerprep, guess what!?

bugigio, you posted the problem incorrectly!
bigduke, you posted the answer for the correct problem!

the correct one should be

3 < x < 5
A) (3 + x) / 3
B) (5 + x) / x

yeah, the denominator for B) should be "x", not "5"

if the denominator is "5", the answer would definetely be A)

cheers everbody

bugigio
08-31-2004, 10:08 AM
1. if u use 12345, the answer would be 2,4,6,9 so try to use random number for example 13472 or else, then you'll got the number, 9.

So I can use any number of 5 digits... I got the question, I got the answer..!



2. bigduke, could you please explain it more detail?
A) ?
B) Combination of 7 to 3?

Yep,that would help me too. I have still a lot of problems to realise if I am facing a combination or a permutation....



3. i know most of this type of questions is "D" (hmm, based on my experience)
but this one, !!! geez, after i look the problem several times, and try to figure bigduke's answer, i found myself looking the problem in powerprep, guess what!?

bugigio, you posted the problem incorrectly!
bigduke, you posted the answer for the correct problem!

the correct one should be

3 < x < 5
A) (3 + x) / 3
B) (5 + x) / x

yeah, the denominator for B) should be "x", not "5"


if the denominator is "5", the answer would definetely be A)

cheers everbody

I can't believe it!!!! So my reasinoning is good...my sight less....
:p :p

Thanks mate!

bigduke
08-31-2004, 10:40 AM
ok chaps lets do the second one like we used to do it in the stone age :D

the combinations for both situations are {3 3 1}, {3 1 3}, {1 3 3}, {3 2 2}, {2 3 2}, {2 2 3}

Now does it make sense?

Also for the divisibility rule for 9, the sum of the dividend's digits must equal 9, therefore a number divisible by 9 must also have its mirror image divisible by 9.

ta da :D

uk007
08-31-2004, 11:51 AM
(1) by divisiblity rule any number subtracted with its reflection will be divisible by 9
because if you write say a 5 digit number it could be as below
number = 10000a+1000b+100c+10d+1e
reflection is =10000e+1000d+100c+10b+a
number -reflection=9999a+990b-0c-990d-9999e
so we know that any number (be any number of digit) subtracted by reflection will be divisible by 9
(2) A and B are not equal

(3)if 3<x<5
then
(3+x)/3 is always greater than (5+x)/5

as (3+x)/3 -(5+x)/5 = x/3-x/5=2x/15 if x>0 2x/15> 0
since x lies between 3 and 5 2x/15 is always positive
hence A>B

uk007
09-02-2004, 11:00 AM
i take back my answer for 2nd ,i read it wrongly.A and B are equal

pami
09-02-2004, 02:59 PM
hey guys,

i found the second question quite interesting. please can anyone explain what are the 6 combinations of the choice B?

dustino
09-03-2004, 04:29 AM
been lurking for awhile, figured id register to see if i can help clear up some confusion on problem 2 :)

the simple answer for the problem can be found in this way....

with 1$ for every person (of 3), what are the most 'extreme' ways to hand out the money? 1, 1, 5 for being lopsided, 2, 2, 3 for being even. the trick here is that there is no possible way to divy out the cash so that one person doesnt have atleast 3$, thus, they are essentially the same question.

the sets that bigduke wrote down were for sets WITH 3$ for atleast one person, however the problem states that there should be at MINIMUM 3$.

no less, here's the simple answer of all the states rewritten out in long-hand form (notice the nice cyclic nature of the problem)

5, 1, 1

4, 2, 1
4, 1, 2

3, 3, 1
3, 2, 2
3, 1, 3

2, 4, 1
2, 3, 2
2, 2, 3
2, 1, 4

1, 5, 1
1, 4, 2
1, 3, 3
1, 2, 4
1, 1, 5

i might consider this a Counting problem (if you were asked how many combinations there were), and as ive heard, a "Triangle" problem. it's called this because for the first series has a cardinality of 1, the second 2, ... to nth series with n, so...

*
**
***
****
*****

we have n = 5 (you cant have 6 because it would leave 1 person with 0$), the formula turns out to be

F(x) = (n * (n + 1)) / 2 = (5 * 6) / 2 = 30 / 2 = 15

pami
09-03-2004, 12:12 PM
exactly dustino, thats what i was contempleting over, there is no other way u can divide the cash unless u use AT LEAST 3$ as one of the options, so the choices become essentially the same. no need to make any further calculations.

pami