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Sub 505 Level|   Number Properties|                     
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ChenggongMAS
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Bunuel
ChenggongMAS
If a is a positive integer, and if the units digit of a^2 is 9 and the units digit of (a+1)^2 is 4, what is the units digit of (a+2)^2?

A) 1
B) 3
C) 5
D) 6
C) 14

I guess I am just not reading this properly. I don't understand what they mean by units digit...

1234.567

1 - THOUSANDS
2 - HUNDREDS
3 - TENS
4 - UNITS
. - decimal point
5 - TENTHS
6 - HUNDREDTHS
7 - THOUSANDTHS

So th units digit is the digit to the left of the decimal point or in integer it's the rightmost digit. For example: the units digit of 1.2 is 1 and the units digit of 13 is 3.

Back to the original question.
If a is a positive integer, and if the units digit of a^2 is 9 and the units digit of (a+1)^2 is 4, what is the units digit of (a+2)^2?

The units digit of a^2 is 9 --> the units digit of a itself is either 3 or 7 (3^2=9 and 7^2=49);
The units digit of (a+1)^2 is 4 --> the units digit of a+1 is either 2 or 8 (2^2=4 and 8^2=64), so the the units digit of a itself is either 2-1=1 or 8-1=7;

To satisfy both conditions the units digit of a must be 7. Now, a+2 will have the units digit equal to 7+2=9, thus the units digit of (a+2)^2, will be 1 (9^2=81).

Answer: A.

Check Number Theory chapter of Math Book for more: math-number-theory-88376.html

I was wondering if there's any algebraic soln to this question.
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Sachin9
Bunuel

I was wondering if there's any algebraic soln to this question.

1. Units digit of a^2 is 9.

2. (a+1)^2 = a^2 + 2a + 1 .....UD of (a^2) + UD of (2a) + 1 = 4 ....9+1 +UD(2a)=4 .....10+UD(2a) = 4...therefore, a = 2 or 7

Based on 1 and 2, a can't be 2, so it has to be 7. We can calculate (a+2)^2

(a+2)^2 = a^2 + 4a + 4 = a^2 + 2(2a) +4 = UD(a^2)+2(UD of 2a) +4 ..this gives you units digit of 1...and thats the answer.
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Hi Bunuel, This one too is tagged as 'hard' in GMATPrep. While it is marked as sub 600 here. Thanks!
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MensaNumber
Hi Bunuel, This one too is tagged as 'hard' in GMATPrep. While it is marked as sub 600 here. Thanks!

You are right but the difficulty level here is based on percentage of users who answered the question correctly/incorrectly: 89% of the users answered this question correctly. Hence the tag.

Does this make sense?
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Bunuel, Thanks for your reply. Yup, it does!

However, I have not seen this logic hold true in every case. What are the percentage ranges for sub600, 600-700, +700 etc? This will help me point out incorrect tags if any so as to improve this forum.

And quite frankly I didnt find this question that easy. But cannot argue against the statistics unless those 101 users somehow were not representative of an average test taker. Moreover, I have heard GMAC also categorizes questions based on how many test takers got it right/wrong. With hundreds of thousands taking the gmat each year they are likely to have bigger data.

I am just trying to understand tagging here. Thanks for your understanding.
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MensaNumber
Bunuel, Thanks for your reply. Yup, it does!

However, I have not seen this logic hold true in every case. What are the percentage ranges for sub600, 600-700, +700 etc? This will help me point out incorrect tags if any so as to improve this forum.

And quite frankly I didnt find this question that easy. But cannot argue against the statistics unless those 101 users somehow were not representative of an average test taker. Moreover, I have hard GMAC also categorizes questions based on how many test takers got it right/wrong. With hundreds of thousands taking the gmat each year they are likely to have bigger data.

I am just trying to understand tagging here. Thanks for your understanding.

Well, you can judge the difficulty level of a question based on the statistics and not on the tags. I agree that GMAC has larger data and their stats might be more representative. Having said that I must add that still the difficulty level is quite subjective issue.
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Gentlemen,

Good afternoon.
It´s my first time in the Forum - I am glad to see such a nice resource! :o


Question: How can I be sure that if the units digit of (a^2 ) = 9 , for sure the units digit of "a" must be 3 or 7 ?

I have followed the answer by expanding the equations and adding the units digits, which I did too, but took quite a longer time.

My first thought when I saw the quation was this " units digit of "a" must e 7 or 9 " approach, however it just sounded in my mind like good a guess - how can I be sure that no other number squared from 0 to infinite will result in a number with 9 ,( or x, or y) in the units digit ? What theory am I missing, guys?

Thank you and luck to all!
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itabra
Gentlemen,

Good afternoon.
It´s my first time in the Forum - I am glad to see such a nice resource! :o


Question: How can I be sure that if the units digit of (a^2 ) = 9 , for sure the units digit of "a" must be 3 or 7 ?

I have followed the answer by expanding the equations and adding the units digits, which I did too, but took quite a longer time.

My first thought when I saw the quation was this " units digit of "a" must e 7 or 9 " approach, however it just sounded in my mind like good a guess - how can I be sure that no other number squared from 0 to infinite will result in a number with 9 ,( or x, or y) in the units digit ? What theory am I missing, guys?

Thank you and luck to all!

If x is an integer to get the units digit of x^2 the only thing we need to know is the units digit of x itself. There are ten digits, so we can have only the following cases:

0^2 = 0
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81

As you can see only if an integer ends with 3 or 9 its square will have the units digit of 9.
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ChenggongMAS
If a is a positive integer, and if the units digit of a^2 is 9 and the units digit of (a+1)^2 is 4, what is the units digit of (a+2)^2?

A. 1
B. 3
C. 5
D. 6
C. 14

I guess I am just not reading this properly. I don't understand what they mean by units digit...

\(a^2=9\) ----> only possible if unit digit is 3 or 7 (\(3^2=9 ; 7^2=49\))
\((a+1)^2=4\); means that a is 7 because (7+1) is 8 and \(8^2=64\) (Unit digit is 4)
now a+2 = 7+2 =9
\(9^2= 81\) (unit digit is 1)
ANSWER is A
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ChenggongMAS
If a is a positive integer, and if the units digit of a^2 is 9 and the units digit of (a+1)^2 is 4, what is the units digit of (a+2)^2?

A. 1
B. 3
C. 5
D. 6
C. 14

Since the units digit of a^2 is 9, the units digit of a is either 3 or 7. However, since the units digit of (a+1)^2 is 4, we see that the units digit of a must equal 7, since then the units digit of a + 1 is 8 and 8^2 = 64 (had the units digit of a been 3, then the units digit of a + 1 would have been 4, but 4^2 = 16). Thus, the units digit of a + 2 is 9, and since 9^2 = 81, the units digit of (a + 2)^2 is 1.

Answer: A
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ChenggongMAS
If a is a positive integer, and if the units digit of a^2 is 9 and the units digit of (a+1)^2 is 4, what is the units digit of (a+2)^2?

A. 1
B. 3
C. 5
D. 6
C. 14

I guess I am just not reading this properly. I don't understand what they mean by units digit...


Since, units digit of \(a^2\) is \(9\) : a can be 3 or 7

Since, units digit of \((a+1)^2\) is \(4\) : a must be 7

Thus, \((a+2)^2 = (7+2)^2 = 81\) , so units digit is 1

Answer must be (A) 1
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Another possible solution without the substitution based on a logical equation:
a^2= x9 (where x - other place values, tenths, hundredths and so on)
(a + 1)^2 = a^2 + 2a + 1 = x4 ,so x9 + 1 + 2a = x0 + 2a = x4
(a + 2)^2 = a^2 + 4a + 4 = x9 +4 + 4а = x3 + 8 = x1 (result of logical equation compared to the above one, with 2а)
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ChenggongMAS
If a is a positive integer, and if the units digit of a^2 is 9 and the units digit of (a+1)^2 is 4, what is the units digit of (a+2)^2?

A. 1
B. 3
C. 5
D. 6
C. 14

I guess I am just not reading this properly. I don't understand what they mean by units digit...

My Approach. 3 and 7 both give unit digit as 9 when squared. but 4 and 8 (a+1 basically) gives 6 and 4 when squared. Means 7 is a possible value. And when 9 is squared it is 81. Hence 1.
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Bunuel

How are you always correct? I am just asking out of curiosity?

Is there any post in GC where you have done any mistake while posting an answer?

Btw amazing man. I wish I had a brain like you.
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It’s simple ,

Think of an integer whose square will have 9 at units place
Either 3 or 7 => any integer having 3 or 7 at units place

Now , check the given part ( integer + 1)^2 is 4 => so , the integer must be 7 because 8^2 is having 4 at units place


Therefore ,( 7+2 )^2= 9^2 => 1 will be at units place .

Answer is A

Posted from my mobile device
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­Slightly trickier variation of a units digit problem:

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JeffTargetTestPrep

I reached the same answer with the following strategy, can you please confirm if this method is valid ?
since a^2 =9 [equation 1], (a+1)^2=4 [equation 2], (a+2)^2=? [equation 4];

from 2: [ a^2 + 2 (a*1) +(1)^2] =4,
from 1 into 2:
[9+ 2a+1] =4
10 + 2a=4;
2a=-6
a=-3 [equation 3]
from 3 into 4
(a+2)^2 =(-3+2)^2=(-1)^2=1
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