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
number properties (720+ level question)
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- kvcpk
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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.
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.
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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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.
Other number that satisfies:18, (18+2)^2 = 400 units digit 0.
Option A.
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- kvcpk
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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!!
(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!!
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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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
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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yes Frank, this comes to be 720+ or level 7 question in Gmat Hacks challenge set for number properties;frank1 wrote:It should be A
initially i thought there was some trap as it was labelled 720+
any way is it?
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
- ikaplan
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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
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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yup i agreeNight 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
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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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.
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.