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# If a,b,c,d,e and f are distinct positive integers and

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If a,b,c,d,e and f are distinct positive integers and [#permalink]

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27 Sep 2006, 03:20
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If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?

(1) a+b+c+d+e+f < 22
(2) b=4 and f=2

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27 Sep 2006, 07:42
Probably would have guessed A on this one on test day.

Spent over 5 mins trying to find a solution.

Basically, all I came up with is a, b, c, d, e, f should be {1, 2, 3, 4, 5 , 6} although not necessarily in that order.

No matter what combination I try to plug in the equation it won't balance.

(E)??

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27 Sep 2006, 22:03

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27 Sep 2006, 23:22
It gets pretty easy when both the statements are used.

If we take the numbers as 1, 2 ,3 ,4 ,5 ,6. ( Dont know if 0 can be a possibility )

10ac+bc= 100d+10e+f
or , c ( 10a + b) = 100d + 10e + f { putting the values of b and f )
or , c ( 10a + 4) = 100d + 10e + 2 {LHS will have 2 in the units place.
In order to match this the value of c
has to be 3 , because 3 * 4 = 12 }
or, 3 (10a + 4) = 100d + 10e + 2
or, 30a + 12 = 100d + 10e + 2
or, 30a + 10 = 100d + 10e
or, 3a + 1 = 10d + e {Now put a = 5 , d = 1 , and e = 6 }

This can be solved. So a = 5

However, my doubt is can't we come to the same conclusion without knowing the values of b and f ?

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27 Sep 2006, 23:27
If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?

(1) a+b+c+d+e+f < 22
(2) b=4 and f=2

So I got soooo bored of writing applications that I had to find something to do with my brain. That is how I am back here. btw.. hope everyone's prep is going on well.

ok here is my attempt.

I satrted with statement B (looks like it has more info than A)
so if (10a+4)c = 100d+10e+2
=> c has to be equal to 3 (you multiply a number ending in 4 with a number ending in 2 the multiple has to be 3.
this quick step tells me b = 4 ; c = 3 & f =2.

but a, d,e could be any numbers

no statement A: this condition gives many possibilities.

combining
a+d+e < 13 (22-4-3-2).....(i)
also 30a+12 = 100d+10e+2
=> 3a +1 = 10d +e.....(ii)
case 1: d= 1
then a+e< 12
only possible value of a that will give each number as a distinct integer is 4

case 2: d=2
then a+e<11
no possible values

case3: d=3
a+e<10
no possible values.

Thus C.

Anyone else tried this? Any other ideas?

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28 Sep 2006, 00:28
If the first statement is a+b+c+d+e+f <=22 then A will be the answer.

But with the given conditions as it is the answer would be E.

Hey kevin where r u?
_________________

Last edited by cicerone on 25 Sep 2008, 01:14, edited 1 time in total.

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28 Sep 2006, 04:02
2times wrote:
If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?

(1) a+b+c+d+e+f < 22
(2) b=4 and f=2

So I got soooo bored of writing applications that I had to find something to do with my brain. That is how I am back here. btw.. hope everyone's prep is going on well.

ok here is my attempt.

I satrted with statement B (looks like it has more info than A)
so if (10a+4)c = 100d+10e+2
=> c has to be equal to 3 (you multiply a number ending in 4 with a number ending in 2 the multiple has to be 3.
this quick step tells me b = 4 ; c = 3 & f =2.

but a, d,e could be any numbers

no statement A: this condition gives many possibilities.

combining
a+d+e < 13 (22-4-3-2).....(i)
also 30a+12 = 100d+10e+2
=> 3a +1 = 10d +e.....(ii)
case 1: d= 1
then a+e< 12
only possible value of a that will give each number as a distinct integer is 4

case 2: d=2
then a+e<11
no possible values

case3: d=3
a+e<10
no possible values.

Thus C.

Anyone else tried this? Any other ideas?

Excellent approach, but were you too quick to dismiss A?

if a+b+c+...+f< 22, then (a,b,c,d,e,f) must be a permutation of the first six positive integers.

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28 Sep 2006, 07:22
kevincan wrote:
2times wrote:
If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?

(1) a+b+c+d+e+f < 22
(2) b=4 and f=2

So I got soooo bored of writing applications that I had to find something to do with my brain. That is how I am back here. btw.. hope everyone's prep is going on well.

ok here is my attempt.

I satrted with statement B (looks like it has more info than A)
so if (10a+4)c = 100d+10e+2
=> c has to be equal to 3 (you multiply a number ending in 4 with a number ending in 2 the multiple has to be 3.
this quick step tells me b = 4 ; c = 3 & f =2.

but a, d,e could be any numbers

no statement A: this condition gives many possibilities.

combining
a+d+e < 13 (22-4-3-2).....(i)
also 30a+12 = 100d+10e+2
=> 3a +1 = 10d +e.....(ii)
case 1: d= 1
then a+e< 12
only possible value of a that will give each number as a distinct integer is 4

case 2: d=2
then a+e<11
no possible values

case3: d=3
a+e<10
no possible values.

Thus C.

Anyone else tried this? Any other ideas?

Excellent approach, but were you too quick to dismiss A?

if a+b+c+...+f< 22, then (a,b,c,d,e,f) must be a permutation of the first six positive integers.

Intuitively, I know that (A) is sufficient, just can't seem to come up with the right combination in under two minutes. Gosh, the GMAT seems to be getting harder all the time

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28 Sep 2006, 08:03
Keep in mind that this is NOT a GMAT question, so don't estimate the difficulty of the exam by the questions I post. GMATPrep is by far the best indicator. These questions are just for fun and meant to help us to look for creative solutions.

If a+b+c+d+e+f<22 and each is a distinct positive integer, it follows that (a,b,c,d,e,f) is a permutation of (1,2,3,4,5,6).

We are told that 10ac+bc=100d+10e+f

Let's write it as:

.ab
x.c
___
def

Focussing on the units digits, it's clear that none of them can be 1 or 5.
Could b and c be 2 and 3 or vice versa? If that case, f would be 6, the two digit number de would simply be ac, which is not possible since the six numbers have to be distinct. This makes it clear that bc>9, since we NEED a carry-over. Only 3 and 4 will do! I have to start a class! Can somebody continue?

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28 Sep 2006, 11:03
Hey kevin, i think the above equation will not have a solution if all of them are distinct and the solution set is 1,2,3,4,5,6

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28 Sep 2006, 14:46
kevincan wrote:
Keep in mind that this is NOT a GMAT question, so don't estimate the difficulty of the exam by the questions I post. GMATPrep is by far the best indicator. These questions are just for fun and meant to help us to look for creative solutions.

If a+b+c+d+e+f<22 and each is a distinct positive integer, it follows that (a,b,c,d,e,f) is a permutation of (1,2,3,4,5,6).

We are told that 10ac+bc=100d+10e+f

Let's write it as:

.ab
x.c
___
def

Focussing on the units digits, it's clear that none of them can be 1 or 5.
Could b and c be 2 and 3 or vice versa? If that case, f would be 6, the two digit number de would simply be ac, which is not possible since the six numbers have to be distinct. This makes it clear that bc>9, since we NEED a carry-over. Only 3 and 4 will do! I have to start a class! Can somebody continue?

so if b and c are 3 and 4 (or vice versa) f would be 2

Let's see if (b,c)=(3,4)

.a3
X 4
----
de2

The two digit number de is 4*a+1- a would have to be 5 or 6, but in each case d would be 2- not permissible

.a4
X.3
----

de2

This time, the two digit number de is 3*a+1- a must be 5, so de is 16

54*3=162

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28 Sep 2006, 14:46
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