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# DS: t divided by 7

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Senior Manager
Joined: 19 Mar 2008
Posts: 346
DS: t divided by 7 [#permalink]

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05 Sep 2008, 03:57
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VP
Joined: 21 Jul 2006
Posts: 1463
Re: DS: t divided by 7 [#permalink]

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05 Sep 2008, 05:26

From the question, we are asked what's the remainder when (t+3)(t+2) is divided by 7:

(1) If t=6, then the remainder will be 2. If t=20, then the remainder will be 4. Not Suff.

(2) When $$t^2$$ is divided by 7, the result will be the same as t divided by 7. whether t= 1, 8, 15, or even 22, the remainder will always be 5.

VP
Joined: 17 Jun 2008
Posts: 1482
Re: DS: t divided by 7 [#permalink]

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05 Sep 2008, 05:45

Stmt1 says that t = 7x + 6 for any integer x.

Thus, t^2 + 5t + 6 will be converted to an expression that when divided by 7 will give a remainder of 2. Hence, sufficient.

Stmt 2: Again using the same logic, the remainder will be different for different values of 7. Hence, insufficient.
Manager
Joined: 20 May 2008
Posts: 55
Re: DS: t divided by 7 [#permalink]

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05 Sep 2008, 06:25
The same set of numbers satify stmt 1 and Stm 2? t= 6, t=13 etc.

So Stmt 1 and Stmt 2 alone are sufficient. Ans should be D.

Whats the OA?
Manager
Joined: 20 May 2008
Posts: 55
Re: DS: t divided by 7 [#permalink]

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05 Sep 2008, 06:32
Just found out the mistake that I was making. t= 6, t = 13 satisfy both stmts but there are some additional values that satify stmt 2 (e.g. t= 1,t=8) and so we dont get the same remainder for all values of t so stmt 2 is out. And should be A
VP
Joined: 21 Jul 2006
Posts: 1463
Re: DS: t divided by 7 [#permalink]

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05 Sep 2008, 06:37
LoyalWater wrote:
Just found out the mistake that I was making. t= 6, t = 13 satisfy both stmts but there are some additional values that satify stmt 2 (e.g. t= 1,t=8) and so we dont get the same remainder for all values of t so stmt 2 is out. And should be A

Is there a way to avoid listing all the numbers? because what if we miss out on a number that can be crucial to our decision like the mistake that you made? By the way, in statement 1, you will get 2 values if you try t= 6 and 20. As for statement 2, t=1 & 8 will still give you the remainder of 5. Plug them into (t+3)(t+2)/7

is there a generic equation that we can just use and avoid listing down all the numbers? because sometimes we might have to list A LOT of numbers until we can arrive to a confirmed answer. I'm starting to hate listing down numbers. Any other faster approach?
Senior Manager
Joined: 19 Mar 2008
Posts: 346
Re: DS: t divided by 7 [#permalink]

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05 Sep 2008, 07:41
tarek99 wrote:
LoyalWater wrote:
Just found out the mistake that I was making. t= 6, t = 13 satisfy both stmts but there are some additional values that satify stmt 2 (e.g. t= 1,t=8) and so we dont get the same remainder for all values of t so stmt 2 is out. And should be A

Is there a way to avoid listing all the numbers? because what if we miss out on a number that can be crucial to our decision like the mistake that you made? By the way, in statement 1, you will get 2 values if you try t= 6 and 20. As for statement 2, t=1 & 8 will still give you the remainder of 5. Plug them into (t+3)(t+2)/7

is there a generic equation that we can just use and avoid listing down all the numbers? because sometimes we might have to list A LOT of numbers until we can arrive to a confirmed answer. I'm starting to hate listing down numbers. Any other faster approach?

Yes.

(1) t = 7k + 6 (where k is integer)
t^2 + 5t +6
= 49k^2 + 14k + 36 + 35k + 30 + 6
= 49k^2 + 14k + 72
(the first two terms are multiple of 7)
so the reminder is equal to the remainder when 72 divided by 7, i.e. 2
(1) is sufficent

(2) t^2 = 7i +1 (where k is integer)
t^2 + 5t +6
= 7i + 1 + 5t + 6
= 7i + 7 + 5t
(the first two terms are multiple of 7)
so the reminder is equal to the remainder when 5t divided by 7

Now we try to find out the remainder when 5t divided by 7:
when i = 0, t^2 = 1, t = 1, remainder of t when divided by 7 = 1
when i = 5, t^2 = 36, t = 6, remainder of t when divided by 7 = 6
so, it give different results.
(2) alone is not suff.

Ans is A
Intern
Joined: 03 Sep 2008
Posts: 22
Re: DS: t divided by 7 [#permalink]

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05 Sep 2008, 07:46
When dealing with remainders you only have to check the numbers that are in the set of possible remainders.
In this problem that set is {0, 1, 2, 3, 4, 5, 6}.

Since 6^2/7 has a remainder of 1, we know all numbers that have a remainder of 6 will have a remainder of 1 when squared.

Since 1^2/7 and 6^2/7 both have remainders of 1, we know that merely knowing the remainder of t^2/7 is not sufficient to determine the remainder of t/7

VP
Joined: 21 Jul 2006
Posts: 1463
Re: DS: t divided by 7 [#permalink]

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Updated on: 07 Sep 2008, 06:34
judokan wrote:
tarek99 wrote:
LoyalWater wrote:
Just found out the mistake that I was making. t= 6, t = 13 satisfy both stmts but there are some additional values that satify stmt 2 (e.g. t= 1,t=8) and so we dont get the same remainder for all values of t so stmt 2 is out. And should be A

Is there a way to avoid listing all the numbers? because what if we miss out on a number that can be crucial to our decision like the mistake that you made? By the way, in statement 1, you will get 2 values if you try t= 6 and 20. As for statement 2, t=1 & 8 will still give you the remainder of 5. Plug them into (t+3)(t+2)/7

is there a generic equation that we can just use and avoid listing down all the numbers? because sometimes we might have to list A LOT of numbers until we can arrive to a confirmed answer. I'm starting to hate listing down numbers. Any other faster approach?

Yes.

(1) t = 7k + 6 (where k is integer)
t^2 + 5t +6
= 49k^2 + 14k + 36 + 35k + 30 + 6
= 49k^2 + 14k + 72
(the first two terms are multiple of 7)
so the reminder is equal to the remainder when 72 divided by 7, i.e. 2
(1) is sufficent

(2) t^2 = 7i +1 (where k is integer)
t^2 + 5t +6
= 7i + 1 + 5t + 6
= 7i + 7 + 5t
(the first two terms are multiple of 7)
so the reminder is equal to the remainder when 5t divided by 7

Now we try to find out the remainder when 5t divided by 7:
when i = 0, t^2 = 1, t = 1, remainder of t when divided by 7 = 1
when i = 5, t^2 = 36, t = 6, remainder of t when divided by 7 = 6
so, it give different results.
(2) alone is not suff.

Ans is A

I have a problem with your statement 2. You provided an equation of $$t^2 = 7i +1$$. So it means:

$$7i+1 + 5sqrt(7i+1) + 6$$ so we don't have 2 variables anymore, but just one because i plugged in your $$t^2 = 7i +1$$ into $$t^2 + 5t + 6$$, which finally means $$7i + 5sqrt(7i+1) + 7$$, so our remainder for statement 2 should be zero because our constant 7 is divisible by 7. what's wrong with that?

Originally posted by tarek99 on 05 Sep 2008, 10:11.
Last edited by tarek99 on 07 Sep 2008, 06:34, edited 1 time in total.
VP
Joined: 21 Jul 2006
Posts: 1463
Re: DS: t divided by 7 [#permalink]

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05 Sep 2008, 12:10
Ok, here's something else that I did, but I still can't tell where I went wrong. Would anyone please show me what I did wrong?

First of all, $$t^2 + 5t + 6$$ can be turned to $$(t+2)(t+3)$$

(1) $$t = 7k + 6$$, so:

$$(7k + 6 + 2) (7k +6 +3) = (7k + 8) (7k + 3) = 49k^2 + 119k + 72$$

Our constant here is 72, which is NOT divisible by 7, so we will have a remainder of 2.

(2) $$t^2 = 7k + 1$$ is the same as $$t = 7k + 1$$, so:

$$(7k + 1 + 2) (7k + 1 + 3) = (7k+3) (7k + 4) = 49k^2 + 49k + 12$$, so:

our constant here is 12, which means that when you divide it by 7, the remainder should be 5. So what's wrong with this approach? I would appreciate it if someone can help me with this problem. regards,
Senior Manager
Joined: 19 Mar 2008
Posts: 346
Re: DS: t divided by 7 [#permalink]

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05 Sep 2008, 20:57
is there a generic equation that we can just use and avoid listing down all the numbers? because sometimes we might have to list A LOT of numbers until we can arrive to a confirmed answer. I'm starting to hate listing down numbers. Any other faster approach?[/quote]

Yes.

(1) t = 7k + 6 (where k is integer)
t^2 + 5t +6
= 49k^2 + 14k + 36 + 35k + 30 + 6
= 49k^2 + 14k + 72
(the first two terms are multiple of 7)
so the reminder is equal to the remainder when 72 divided by 7, i.e. 2
(1) is sufficent

(2) t^2 = 7i +1 (where k is integer)
t^2 + 5t +6
= 7i + 1 + 5t + 6
= 7i + 7 + 5t
(the first two terms are multiple of 7)
so the reminder is equal to the remainder when 5t divided by 7

Now we try to find out the remainder when 5t divided by 7:
when i = 0, t^2 = 1, t = 1, remainder of t when divided by 7 = 1
when i = 5, t^2 = 36, t = 6, remainder of t when divided by 7 = 6
so, it give different results.
(2) alone is not suff.

Ans is A[/quote]

I have a problem with your statement 2. You provided an equation of $$t^2 = 7i +1$$. So it means:

$$7i+1 + 5sqrt(7i+1) + 6$$ so we don't have 2 variables anymore, but just one because i plugged in your $$t^2 = 7i +1$$ into $$t^2 + 5t + 6$$, which finally means $$7i + 5sqrt(7i+1) + 7$$, so our remainder for statement 2 should be zero because our constant 7 is visible by 7. what's wrong with that?[/quote]

Hi, the reason we have to to find out the possible number of t is that we do not know the remainder when t is divided by 7.

Can you tell the remainder when (7i+1)^(0.5) divided by 7?

On the other hand, why t^2 + 6 = 7i + 7, which is always divisible by 7, therefore remainder is 0.

So we need to find out the remainder when (7i+1)^(0.5) divided by 7 in order to find out the remainder when t^2 + 5t +6 is divided by 7.

So the next step is to insert diffenert possible value of i, to find t, to see whether it gives consistent answer.

Since t is a positive integer.
when i = 0, t^2 = 1, t = 1, remainder of t when divided by 7 = 1
when i = 5, t^2 = 36, t = 6, remainder of t when divided by 7 = 6

So we only have variable answer, so not suff to solve the problem
Senior Manager
Joined: 19 Mar 2008
Posts: 346
Re: DS: t divided by 7 [#permalink]

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05 Sep 2008, 21:01
tarek99 wrote:
Ok, here's something else that I did, but I still can't tell where I went wrong. Would anyone please show me what I did wrong?

First of all, $$t^2 + 5t + 6$$ can be turned to $$(t+2)(t+3)$$

(1) $$t = 7k + 6$$, so:

$$(7k + 6 + 2) (7k +6 +3) = (7k + 8) (7k + 3) = 49k^2 + 119k + 72$$

Our constant here is 72, which is NOT divisible by 7, so we will have a remainder of 2.

(2) $$t^2 = 7k + 1$$ is the same as $$t = 7k + 1$$, so:

$$(7k + 1 + 2) (7k + 1 + 3) = (7k+3) (7k + 4) = 49k^2 + 49k + 12$$, so:

our constant here is 12, which means that when you divide it by 7, the remainder should be 5. So what's wrong with this approach? I would appreciate it if someone can help me with this problem. regards,

The mistake is
$$t^2 = 7k + 1$$ is NOT the same as $$t = 7k + 1$$

if $$t^2 = 7*5 + 1$$
$$t = 6$$ when divided by 7, the remainder is 6, or $$t = 7i + 6$$

if $$t^2 = 7*0 + 1$$
$$t = 1$$ when divided by 7, the remainder is 1, or $$t = 7i + 1$$
VP
Joined: 17 Jun 2008
Posts: 1295
Re: DS: t divided by 7 [#permalink]

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05 Sep 2008, 21:20
judokan wrote:

(1)substitute 7p+6 we will get remainder 2 indepoendent of value of t
hence r=2 SUFFI

(2) remainder of t^2 is known but no of t hence INSUFFI

IMO A
_________________

cheers
Its Now Or Never

VP
Joined: 21 Jul 2006
Posts: 1463
Re: DS: t divided by 7 [#permalink]

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07 Sep 2008, 06:46
judokan wrote:
tarek99 wrote:
Ok, here's something else that I did, but I still can't tell where I went wrong. Would anyone please show me what I did wrong?

First of all, $$t^2 + 5t + 6$$ can be turned to $$(t+2)(t+3)$$

(1) $$t = 7k + 6$$, so:

$$(7k + 6 + 2) (7k +6 +3) = (7k + 8) (7k + 3) = 49k^2 + 119k + 72$$

Our constant here is 72, which is NOT divisible by 7, so we will have a remainder of 2.

(2) $$t^2 = 7k + 1$$ is the same as $$t = 7k + 1$$, so:

$$(7k + 1 + 2) (7k + 1 + 3) = (7k+3) (7k + 4) = 49k^2 + 49k + 12$$, so:

our constant here is 12, which means that when you divide it by 7, the remainder should be 5. So what's wrong with this approach? I would appreciate it if someone can help me with this problem. regards,

The mistake is
$$t^2 = 7k + 1$$ is NOT the same as $$t = 7k + 1$$

if $$t^2 = 7*5 + 1$$
$$t = 6$$ when divided by 7, the remainder is 6, or $$t = 7i + 6$$

if $$t^2 = 7*0 + 1$$
$$t = 1$$ when divided by 7, the remainder is 1, or $$t = 7i + 1$$

cool, so even if i was wrong about $$t^2 = 7k + 1$$ being the same as $$t = 7k + 1$$. I still can create an equation, no? Here's statement 2 again:

$$t^2 = 7k + 1$$, so if i plug it into the given equation $$t^2 + 5t + 6$$, i would get:

$$(7k + 1) + 5 sqrt(7k+1) + 6$$, which is equal to $$7k + 5sqrt(7k+1) + 7$$, so what's wrong with this? why aren't we considering the 7 in the last part of the addition to conclude that the remainder will be zero? 7 is the constant in this equation after all. Haven't we been picking the constant of every such formula all the time? why aren't we doing so here this time? That's my main question.
regards,
Senior Manager
Joined: 19 Mar 2008
Posts: 346
Re: DS: t divided by 7 [#permalink]

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07 Sep 2008, 09:29
tarek99 wrote:

cool, so even if i was wrong about $$t^2 = 7k + 1$$ being the same as $$t = 7k + 1$$. I still can create an equation, no? Here's statement 2 again:

$$t^2 = 7k + 1$$, so if i plug it into the given equation $$t^2 + 5t + 6$$, i would get:

$$(7k + 1) + 5 sqrt(7k+1) + 6$$, which is equal to $$7k + 5sqrt(7k+1) + 7$$, so what's wrong with this? why aren't we considering the 7 in the last part of the addition to conclude that the remainder will be zero? 7 is the constant in this equation after all. Haven't we been picking the constant of every such formula all the time? why aren't we doing so here this time? That's my main question.
regards,

: )

is 14 + 7 divisible by 7?
is 16 + 7 divisible by 7?
If we can conclude the whole number is divisible by 7 by pulling out a contant 7, then all number can be divisble by 7.
99 = 92 + 7, then is 99 divisible by 7? the remainder is 0? sure not.

if x = A + B + C, if A and B are both divisible by 7 (i.e. remainder is 0), then we know that the remainder when C is divided by 7, is equal to the remainder when x divided by 7.

If A, B are not divisible by 7, we need to find out the remainder when they are divided by 7, in order to find the remainder of x when divided by 7.

Go back to your equation, very nice, 100% correct.

$$7k + 5sqrt(7k+1) + 7$$

A = 7K
B = 7
A & B are divisible by 7, so remainder is 0

Then we can conclude that the remainder in question is equal to the remainder when $$5sqrt(7k+1)$$ divided by 7.

Is it divisible by 7? yes or no. Clearly, we cannot arrive a certain yes, because we cannot pull out a factor of 7 from it, like this form $$7*5sqrt(7k+1)$$

In order to prove that no certain answer can be reached, we try to fill in some numbers for k.

After filling in the numbers, we can 100% conclude that the term $$5sqrt(7k+1)$$ and therefore $$t^2 + 5t + 6$$ may or maynot be divisible by 7 (varies remainder), depending the actual value of k (i.e. t)
VP
Joined: 21 Jul 2006
Posts: 1463
Re: DS: t divided by 7 [#permalink]

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07 Sep 2008, 10:30
judokan wrote:
tarek99 wrote:

cool, so even if i was wrong about $$t^2 = 7k + 1$$ being the same as $$t = 7k + 1$$. I still can create an equation, no? Here's statement 2 again:

$$t^2 = 7k + 1$$, so if i plug it into the given equation $$t^2 + 5t + 6$$, i would get:

$$(7k + 1) + 5 sqrt(7k+1) + 6$$, which is equal to $$7k + 5sqrt(7k+1) + 7$$, so what's wrong with this? why aren't we considering the 7 in the last part of the addition to conclude that the remainder will be zero? 7 is the constant in this equation after all. Haven't we been picking the constant of every such formula all the time? why aren't we doing so here this time? That's my main question.
regards,

: )

is 14 + 7 divisible by 7?
is 16 + 7 divisible by 7?
If we can conclude the whole number is divisible by 7 by pulling out a contant 7, then all number can be divisble by 7.
99 = 92 + 7, then is 99 divisible by 7? the remainder is 0? sure not.

if x = A + B + C, if A and B are both divisible by 7 (i.e. remainder is 0), then we know that the remainder when C is divided by 7, is equal to the remainder when x divided by 7.

If A, B are not divisible by 7, we need to find out the remainder when they are divided by 7, in order to find the remainder of x when divided by 7.

Go back to your equation, very nice, 100% correct.

$$7k + 5sqrt(7k+1) + 7$$

A = 7K
B = 7
A & B are divisible by 7, so remainder is 0

Then we can conclude that the remainder in question is equal to the remainder when $$5sqrt(7k+1)$$ divided by 7.

Is it divisible by 7? yes or no. Clearly, we cannot arrive a certain yes, because we cannot pull out a factor of 7 from it, like this form $$7*5sqrt(7k+1)$$

In order to prove that no certain answer can be reached, we try to fill in some numbers for k.

After filling in the numbers, we can 100% conclude that the term $$5sqrt(7k+1)$$ and therefore $$t^2 + 5t + 6$$ may or maynot be divisible by 7 (varies remainder), depending the actual value of k (i.e. t)

oohhhh.....i see, i see...now i understand what's going on hear. Thanks a lot for the great explanation. I understand now. Thanks again

Another question I have for you. Let's say that x = a + b + c and we're trying to divide x by 7. Suppose that when a is divided by 7, the remainder is 2; when b is divided by 7, the remainder is 1; and when c is divided by 7, the remainder is 3. does that mean that the overall remainder when x is divided by 7 will be the addition of each of these remainders? so 2+1+3 = 6?
Senior Manager
Joined: 19 Mar 2008
Posts: 346
Re: DS: t divided by 7 [#permalink]

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07 Sep 2008, 10:59
tarek99 wrote:

Another question I have for you. Let's say that x = a + b + c and we're trying to divide x by 7. Suppose that when a is divided by 7, the remainder is 2; when b is divided by 7, the remainder is 1; and when c is divided by 7, the remainder is 3. does that mean that the overall remainder when x is divided by 7 will be the addition of each of these remainders? so 2+1+3 = 6?

Hi tarek, seems you totally understand now.

Yes, the remainder is 6.

To elaborate more, (actually, the method is very similar to what we used above)

Suppose that when a is divided by 7, the remainder is 2; when b is divided by 7, the remainder is 1; and when c is divided by 7, the remainder is 3

then we can let:
a = 7q + 2
b = 7z + 1
c = 7i + 3

then x = 7*(q+z+i) + 6
so the remainder is 6

let's look at one more example:
A = 7Q + 6
B = 7Z + 6
C = 7I + 5
then X = 7*(Q+Z+I) + 17
So remainder is equal to the remainder when 17 divided by 7, i.e. 3

if you still confused, think about
then X = 7*(Q+Z+I) + 7 + 10
then only the term 10 is not divisble by 7
so the remainder is equal to 10 divided by 7, i.e. 3
VP
Joined: 21 Jul 2006
Posts: 1463
Re: DS: t divided by 7 [#permalink]

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07 Sep 2008, 11:19
judokan wrote:
tarek99 wrote:

Another question I have for you. Let's say that x = a + b + c and we're trying to divide x by 7. Suppose that when a is divided by 7, the remainder is 2; when b is divided by 7, the remainder is 1; and when c is divided by 7, the remainder is 3. does that mean that the overall remainder when x is divided by 7 will be the addition of each of these remainders? so 2+1+3 = 6?

Hi tarek, seems you totally understand now.

Yes, the remainder is 6.

To elaborate more, (actually, the method is very similar to what we used above)

Suppose that when a is divided by 7, the remainder is 2; when b is divided by 7, the remainder is 1; and when c is divided by 7, the remainder is 3

then we can let:
a = 7q + 2
b = 7z + 1
c = 7i + 3

then x = 7*(q+z+i) + 6
so the remainder is 6

let's look at one more example:
A = 7Q + 6
B = 7Z + 6
C = 7I + 5
then X = 7*(Q+Z+I) + 17
So remainder is equal to the remainder when 17 divided by 7, i.e. 3

if you still confused, think about
then X = 7*(Q+Z+I) + 7 + 10
then only the term 10 is not divisble by 7
so the remainder is equal to 10 divided by 7, i.e. 3

perfect. well understood. Thanks a lot for your explanation. I totally understand now.
VP
Joined: 21 Jul 2006
Posts: 1463
Re: DS: t divided by 7 [#permalink]

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07 Sep 2008, 20:13
i've also noticed something interesting. For example:

14 = 5 + 9

suppose we're trying to divide 14 by 7. 14 is divisible by 7. However, when you separately divide 5 by 7 and 9 by 7, non of them are divisible by 7. But when the sum of their remainders are a multiple of 7, then the whole number is divisible by 7. For example: 5 divide by 7, the remainder is 5. When you divide 9 by 7, the remainder is 2. Therefore, 5+2 = 7, which is a multiple of 7, so 14 is divisible by 7. Is that a correct way to approach such a situation?
Senior Manager
Joined: 19 Mar 2008
Posts: 346
Re: DS: t divided by 7 [#permalink]

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07 Sep 2008, 21:54
tarek99 wrote:
i've also noticed something interesting. For example:

14 = 5 + 9

suppose we're trying to divide 14 by 7. 14 is divisible by 7. However, when you separately divide 5 by 7 and 9 by 7, non of them are divisible by 7. But when the sum of their remainders are a multiple of 7, then the whole number is divisible by 7. For example: 5 divide by 7, the remainder is 5. When you divide 9 by 7, the remainder is 2. Therefore, 5+2 = 7, which is a multiple of 7, so 14 is divisible by 7. Is that a correct way to approach such a situation?

Yes, this is correct. Good suggestion.

You may also want to think about the following question.

What is the remainder when n is divided by 9? n is a positive integer.

(1) the sum of all digits of n is divisible by 11.

(i forgot statement (2))

Ans is A or E?
Re: DS: t divided by 7   [#permalink] 07 Sep 2008, 21:54

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