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If (C-D)/(C+D) - (C+D)/(C-D) = 0, what is the value of C (1)

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If (C-D)/(C+D) - (C+D)/(C-D) = 0, what is the value of C (1) [#permalink] New post 11 Mar 2007, 06:03
1. If (C-D)/(C+D) - (C+D)/(C-D) = 0, what is the value of C

(1) -4CD = 0
(2) D>0

2. If x and y are +ve integers, (2^Y)X = 176, what is the least possible value of XY?

3. If n and k are +ve integers, what is the value of the unit's digit of (7^(4k+3))(6^n)?

Again, I have no answers for these. :cry:
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 [#permalink] New post 11 Mar 2007, 06:15
I have tried to solve these questions. My answers are:

1. C = 0

2. XY = 44 => 4*11 (X=4 and Y = 11)

3. I have four unit's digits: 2, 4, 6, 8. I don't know which one is correct.
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 [#permalink] New post 11 Mar 2007, 06:55
For question 1. answer should be B

here given c-d/c+d - c+d/c-d =0 will become -4cd/(c+d)x(c-d)=0

Now if we consider one i.e. -4cd=0 we will not reach to conclusion. so its insufficient

but from 2) we know that if d>0 then c should be 0 hence 2) is sufficient hence the answer is B

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 [#permalink] New post 11 Mar 2007, 07:00
For question number 2 answer is 64


2. If x and y are +ve integers, (2^Y)X = 176, what is the least possible value of XY?

Explanation to find out the least value of X*Y, we need to find out the maximum value of 2^Y, which will be divisible by 176. max value is 16 that means Y will 4 and X will be 16

now 4*16=64

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 [#permalink] New post 11 Mar 2007, 07:05
For question number 3 answer is 8

3. If n and k are +ve integers, what is the value of the unit's digit of (7^(4k+3))(6^n)?

this equation can be broken as 7^3*7^4k*6^n

Solving this we will have the unit's digit for 6^n for any value of n is always 6.

Also unit digit of 7^3 is 3, while unit digit for 7^4k for any value of K is 1. you may solve for K by putting various numbers.

So on multiplying these units i.e. 3*1*6 will give unit digit 8

hence 8 is the answer.

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 [#permalink] New post 11 Mar 2007, 10:10
For Q1 Agree with Amandeep
For Q2 (2^y)x=176

16*11--thus y=4 and x=11 xy=44
8*22 thus y=3 and x=22 xy=88
4*44 thus y=2 and x=88 xy=88
2*88 thus y=1 and x=88 xy=176
2^0 *176 thus y=0 and x=176 xy=0

Thus min value is zero
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 [#permalink] New post 11 Mar 2007, 10:16
Oh I missed that... you are right yogesh... xy=0...

thnx for correcting me... it was a silly mistake, which I need to avoid during GMAT.

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 [#permalink] New post 11 Mar 2007, 15:44
Thanks. I misunderstood the Q1, but now I got it. As for Q2, I also missed that y=0 part.

Now waiting for Q3's solution.... :)

Thanks!
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 [#permalink] New post 11 Mar 2007, 18:34
for Q2, it says that x and y are +ve integers. this implies they cant be zero. so for the product to be minimum, we take the max possible value of y which is 4. the value of x, under this case is 11. so the ans is 44.
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 [#permalink] New post 11 Mar 2007, 18:37
for Q3, again n and k are +ve. so take n = k = 1. unit's place of 7^7 is 3. for 6^1 its 6. so the ans is 8.
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 [#permalink] New post 11 Mar 2007, 20:21
yogeshsheth wrote:
For Q1 Agree with Amandeep
For Q2 (2^y)x=176

16*11--thus y=4 and x=11 xy=44
8*22 thus y=3 and x=22 xy=88
4*44 thus y=2 and x=88 xy=88
2*88 thus y=1 and x=88 xy=176
2^0 *176 thus y=0 and x=176 xy=0

Thus min value is zero


lets review the condition
"If x and y are +ve integers, (2^Y)X = 176, what is the least possible value of XY?

y must be greater than 0. thus the answer is x*y=44, where y=4 and x=11.
"
  [#permalink] 11 Mar 2007, 20:21
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