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number properties (720+ level question)

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by Night reader » Mon Dec 13, 2010 8:57 pm
If k is a positive integer, and if the units' digit of k^2 is 4 and the units' digit of (k + 1)^2 is 1, what is the units' digit of (k + 2)^2 ?

(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
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by kvcpk » Mon Dec 13, 2010 9:09 pm
IMO A.

This might be a 720+ problem if we need to solve it arithmetically. If we plugin, it should be simple.

if k^2 has to end in 4, k has to end in either 2 or 8
if (k+1)^2 has to end in 1, k+1 has to end in 1 or 9

Let us say k=2,
k+1 =3 will not end in 1 or 9
hence k cant be 2

Let us say k=8,
k+1 = 9 ends in 9

Hence k+2 = 10
(k+2)^2 = 100 will end in 0

pick A.
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by beat_gmat_09 » Mon Dec 13, 2010 9:10 pm
Numbers that satisfy the conditions have 8 as their units digit, Number+1 has 9 as its units digit which when squared gives 1 at the units digit. Numbers can be - 8 : 8^2 = 64, (8+1)^2 81 units digit 1 (8+2)^2 100 units digit 0.
Other number that satisfies:18, (18+2)^2 = 400 units digit 0.
Option A.
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by kvcpk » Mon Dec 13, 2010 9:15 pm
arithmetically:

(k+1)^2 = k^2 + 2k +1
if k^2 ends in 4, k will end in 2 or 8
hence 2k will end in 4 or 6

representing last digit as LD
LD(k+1)^2 = LD(4)+LD(4)+1 = LD(9)
9 cannot equal 1.

LD(k+1)^2 = LD(4)+LD(6)+1 = LD(1)
Hence satisfied.
Hence k will end in 8.

(K+2)^2 = k^2 + 4k + 4
LD(k+2)^2 = LD(4)+2*LD(6)+4 = LD(4)+LD(2)+4 = LD(0)

Hence last digit is 0.

Hope this helps!!
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by Night reader » Mon Dec 13, 2010 9:40 pm
thank you guys, below is my solution took around 60 sec
k^2=... 4
(k+1)^2= ... 1 <=> k^2+2k+1= ... 1 <=> 2(k^2+2k+1)= ... 2
(k+2)^2= ... ? <=> k^2+4k+4= ... ? <=> ...2+2-...4= ...0
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by frank1 » Tue Dec 14, 2010 12:47 am
It should be A

initially i thought there was some trap as it was labelled 720+

any way is it?
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by Night reader » Tue Dec 14, 2010 1:56 am
frank1 wrote:It should be A

initially i thought there was some trap as it was labelled 720+

any way is it?
yes Frank, this comes to be 720+ or level 7 question in Gmat Hacks challenge set for number properties;

as you, I am also discouraged with many problems in this set- I simply find them easy...
may be test has trained us to be monsters
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by ikaplan » Tue Dec 14, 2010 7:16 am
I listed all digits from 1-9 and squared them- it turned out that both 2 and 8 satisfy the first condition (k^2=4).

Then I plugged in 2 and 8 into the second condition- (k+1)^2=1 and only 8 satisfied the condition (so 2 is out).

Lastly, I plugged in 8 into (k+2)^2 ---> (8+2)^2=100 so

IMO- A is the correct choice
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by frank1 » Tue Dec 14, 2010 7:59 am
Night reader wrote:
yes Frank, this comes to be 720+ or level 7 question in Gmat Hacks challenge set for number properties;

as you, I am also discouraged with many problems in this set- I simply find them easy...
may be test has trained us to be monsters
yup i agree
and may be another thing is in exam scenario or under timed condition,the difficulty of the question depends upon several factors..
i have experience of making error in even simple question when i was running for time or i have very limited time to finish test
or when i was not confident of earlier question....
so i think these are all reasons which makes gmat difficult
otherwise if they allow 4-5 hours to solve thing i feel there will be many 770+ people...
just what i feel

best of luck with you preparation

i agree when i first look at the things question stem seems intimidating as we feel it may require lot of calculations
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by Woozler » Tue Dec 14, 2010 9:22 am
I am mediocre at quant and horrid at number properties, yet I solved it by just plugging in all even numbers (2, 4, 6, 8) and then added one to them. 8 is the only one that worked, therefore (A)

Is this really a 720+ level problem? I wouldn't have been able to solve it if it really were. I score 46-49 on quant sections.
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