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When a number a is divided by 6, the remainder is 3 and when another
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13 Jul 2023, 06:05
13 Jul 2023, 06:05
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Question Stats:
65% (01:48) correct
34% (01:39) wrong based on 23 sessions
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When a number ‘a’ is divided by 6, the remainder is 3 and when another number ‘b’ is divided by 12, the remainder is 9. What is the remainder when a^2 + b^2 is divided by 12?
A. 4
B. 5
C. 6
D. 10
E. None of the above
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE NUMBER PROPERTIES
A. 4
B. 5
C. 6
D. 10
E. None of the above
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE NUMBER PROPERTIES
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Re: When a number a is divided by 6, the remainder is 3 and when another
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14 Jul 2023, 13:06
14 Jul 2023, 13:06
1
Expert Reply
To find the remainder when a^2+b^2 is divided by 12, we need to consider the remainders when a and b are divided by 6 and 12, respectively.
Given that the remainder when a is divided by 6 is 3, we can express a as: a=6k+3, where k is an integer.
Similarly, the remainder when b is divided by 12 is 9, so we can express b as: b=12m+9, where m is an integer.
Now, let's substitute these expressions into a^2+b^2 and simplify:
a2+b2=(6k+3)2+(12m+9)2
=36k^2+36k+9+144m^2+216m+81
=36k^2+144m^2+36k+216m+90
Notice that 36 is divisible by both 6 and 12, so we can ignore those terms. Thus, the remainder when a^2+b^2 is divided by 12 will be the remainder when
36k+216m+90 is divided by 12.
Let's simplify further: 36k+216m+90=12(3k+18m+7)+6
From this expression, we can see that when 36k+216m+90 is divided by 12, the remainder is 6.
Therefore, the correct answer is (C) 6.
Given that the remainder when a is divided by 6 is 3, we can express a as: a=6k+3, where k is an integer.
Similarly, the remainder when b is divided by 12 is 9, so we can express b as: b=12m+9, where m is an integer.
Now, let's substitute these expressions into a^2+b^2 and simplify:
a2+b2=(6k+3)2+(12m+9)2
=36k^2+36k+9+144m^2+216m+81
=36k^2+144m^2+36k+216m+90
Notice that 36 is divisible by both 6 and 12, so we can ignore those terms. Thus, the remainder when a^2+b^2 is divided by 12 will be the remainder when
36k+216m+90 is divided by 12.
Let's simplify further: 36k+216m+90=12(3k+18m+7)+6
From this expression, we can see that when 36k+216m+90 is divided by 12, the remainder is 6.
Therefore, the correct answer is (C) 6.
Re: When a number a is divided by 6, the remainder is 3 and when another
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25 Oct 2023, 05:57
25 Oct 2023, 05:57
1
Just take small values for for a and b, for example I took, a=3 and b=9, so that when a divided 6 remainder is 3 and when b divided 12 remainder is 9.
now a^2+b^2=9+81=90
and when we divide a^2+b^2 which is 90 by 12, we get remainder 6.
so C is the answer.
Is it a good and fast way?
now a^2+b^2=9+81=90
and when we divide a^2+b^2 which is 90 by 12, we get remainder 6.
so C is the answer.
Is it a good and fast way?
