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Algebra.

by govind_raj_76 » Sun Jul 18, 2010 8:38 am
Chris joined IBC consulting. The password given to Chris to access the computer lab is a 4 digit number. The 'n' th digit of the password is the remainder when expression n^2 + 24n + 61 is divided by the expression n+3 after substituting the value of n.

What is the password that Chris got fot the computer lab ?

A. 4567

B. 3456

c. 2311

d. 2341

e. 2345

Please assist.
Govind
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by aloneontheedge » Sun Jul 18, 2010 9:07 am
govind_raj_76 wrote:Chris joined IBC consulting. The password given to Chris to access the computer lab is a 4 digit number. The 'n' th digit of the password is the remainder when expression n^2 + 24n + 61 is divided by the expression n+3 after substituting the value of n.

What is the password that Chris got fot the computer lab ?

A. 4567

B. 3456

c. 2311

d. 2341

e. 2345

Please assist.
E is the answer. tried by hit and trial
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by kmittal82 » Sun Jul 18, 2010 11:50 am
There seems to be no easy way of doing this, so lets jump right in. Firstly, divide the equations and you will see that:

n^2 + 24n + 61 = (n+3)(n+21) - 2

The expression on the right hand side is easier to calculate

So, for the first digit, remainder is [(4x22) - 2 ]/4 = 86/4 = 2. So we know first digit is 2

This eliminates (a) and (b).

Between c, d and e, the 3rd digits are different, so lets see what the 3rd digit should be:

Remainder when (3+3)(3+21) - 2 is divided by 3+3 => [(6x24)-2]/6 -> Remainder of 4, which leaves only d and e. Now for the 4th digit:

Remainder = (7x25 - 2) / 7 -> remainder of 5

So, only [spoiler]option (E)[/spoiler] remains, which should be the right answer. Can we have the OA please?
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by Stuart@KaplanGMAT » Sun Jul 18, 2010 1:32 pm
govind_raj_76 wrote:Chris joined IBC consulting. The password given to Chris to access the computer lab is a 4 digit number. The 'n' th digit of the password is the remainder when expression n^2 + 24n + 61 is divided by the expression n+3 after substituting the value of n.

What is the password that Chris got fot the computer lab ?

A. 4567

B. 3456

c. 2311

d. 2341

e. 2345

Please assist.
Definitely want to use the answer choices to cut down the math.

Before we dive in to calculations, let's look at the 4 digits.

First digit: choices are 4, 3, 2, 2, 2
Second digit: choices are 5, 4, 3, 3, 3
Third digit: choices are 6, 5, 1, 4, 4
Fourth digit: choices are 7, 6, 1, 1, 5

Since the third and fourth digits have more options, we should calculate those first - we'll be able to eliminate more choices.

Let's use the 4th digit.

If n=4, then:

n^2 + 24n + 61 = 16 + 96 + 61 = 173

n+3 = 7

173/7 = 24rem5

Only (E) ends in 5... choose (E)!

If we had gotten a remainder of 1, both (C) and (D) would still be in the running and we'd have calculated the 3rd digit as well (since that's the only difference between the two numbers).
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by jaskaran » Sun Jul 18, 2010 1:58 pm
Nicely put Stuart, thanks! but isnt there a more eloquent way of solving this?
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by Stuart@KaplanGMAT » Sun Jul 18, 2010 2:01 pm
jaskaran wrote:Nicely put Stuart, thanks! but isnt there a more eloquent way of solving this?
Depends how you define "eloquent"; if by eloquent you mean "solving 4 incredibly complicated equations over the course of 8 minutes", then yes, there is.

:mrgreen:
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by jaskaran » Sun Jul 18, 2010 2:09 pm
Stuart Kovinsky wrote:
jaskaran wrote:Nicely put Stuart, thanks! but isnt there a more eloquent way of solving this?
Depends how you define "eloquent"; if by eloquent you mean "solving 4 incredibly complicated equations over the course of 8 minutes", then yes, there is.

:mrgreen:
lol, I was hoping for more like within 2 mins, too optimistic, I suppose!...
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