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10 Jun 2020, 04:00
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The expression \(2a^2 +2b^2\) can be written in the form of \(18x + y\), where a, b, x and y are non-negative integers and \(y < 18\). Is \(|y + 3| = 3\)?
(1) The difference between a and b can be expressed as an even multiple of 3.
(2) b when divided by 3 is an integer.
Are You Up For the Challenge: 700 Level Questions
(1) The difference between a and b can be expressed as an even multiple of 3.
(2) b when divided by 3 is an integer.
Are You Up For the Challenge: 700 Level Questions
yashikaaggarwal
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10 Jun 2020, 05:27
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IMO C
2A^2+2B^2 = 18x+y
Statement 1: The difference between a and b can be expressed as an even multiple of 3.
Therefore A-B = 6,12,18......
Since A,B are nonnegative integers.
Let,
A-B = 6 therefore,
7-1 = 6, 8-2 = 6, 9-3 = 6
The pairs of A and B be (7,1)(8,2)(9,3)(10,4)(11,5)..........
The possible pair of A and B = (7,1)(8,2)(9,3)(10,4)(11,5)....
Putting values in 2A^2+2B^2 = 18x+y
(7,1)
2*7^2+2*1^2 = 18x+y
98+2 = 18x+y
100 = 18x+y
Nearest multiple of 18 to 100 is 90
Therefore X = 5 & y = 10 (not possible)
(8,2)
2*8^2+2*2^2 = 18x+y
128+8 = 18x+y
136 = 18x+y
Nearest multiple of 18 to 136 is 126
Therefore X = 7 & y = 10 (not possible)
(9,3)
2*9^2+2*3^2 = 18x+y
162+18 = 18x+y
180 = 18x+y
Nearest multiple of 18 to 180 is 180
Therefore X = 10 & y = 0 (not possible)
(Not Sufficient)
Statement 2: B when divided by 3 leaves an integer.
But we need at least A's value to determine Y(not sufficient)
Together:
(9,3)
2*9^2+2*3^2 = 18x+y
162+18 = 18x+y
180 = 18x+y
Nearest multiple of 18 to 180 is 180
Therefore X = 10 & y = 0
(12,6)
2*12^2+2*6^2 = 18x+y
288+72 = 18x+y
360 = 18x+y
Nearest multiple of 18 to 360 is 360
Therefore X = 20 & y = 0
(21,15)
2*21^2+2*15^2 = 18x+y
882+450 = 18x+y
1332 = 18x+y
Nearest multiple of 18 to 1332 is 1332
Therefore X = 74 & y = 0
Therefore for every B which is a multiple of 3 gives out y value as 0.
|y+3| = 3
|0+3| = 3
3 = 3
(Sufficient)
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2A^2+2B^2 = 18x+y
Statement 1: The difference between a and b can be expressed as an even multiple of 3.
Therefore A-B = 6,12,18......
Since A,B are nonnegative integers.
Let,
A-B = 6 therefore,
7-1 = 6, 8-2 = 6, 9-3 = 6
The pairs of A and B be (7,1)(8,2)(9,3)(10,4)(11,5)..........
The possible pair of A and B = (7,1)(8,2)(9,3)(10,4)(11,5)....
Putting values in 2A^2+2B^2 = 18x+y
(7,1)
2*7^2+2*1^2 = 18x+y
98+2 = 18x+y
100 = 18x+y
Nearest multiple of 18 to 100 is 90
Therefore X = 5 & y = 10 (not possible)
(8,2)
2*8^2+2*2^2 = 18x+y
128+8 = 18x+y
136 = 18x+y
Nearest multiple of 18 to 136 is 126
Therefore X = 7 & y = 10 (not possible)
(9,3)
2*9^2+2*3^2 = 18x+y
162+18 = 18x+y
180 = 18x+y
Nearest multiple of 18 to 180 is 180
Therefore X = 10 & y = 0 (not possible)
(Not Sufficient)
Statement 2: B when divided by 3 leaves an integer.
But we need at least A's value to determine Y(not sufficient)
Together:
(9,3)
2*9^2+2*3^2 = 18x+y
162+18 = 18x+y
180 = 18x+y
Nearest multiple of 18 to 180 is 180
Therefore X = 10 & y = 0
(12,6)
2*12^2+2*6^2 = 18x+y
288+72 = 18x+y
360 = 18x+y
Nearest multiple of 18 to 360 is 360
Therefore X = 20 & y = 0
(21,15)
2*21^2+2*15^2 = 18x+y
882+450 = 18x+y
1332 = 18x+y
Nearest multiple of 18 to 1332 is 1332
Therefore X = 74 & y = 0
Therefore for every B which is a multiple of 3 gives out y value as 0.
|y+3| = 3
|0+3| = 3
3 = 3
(Sufficient)
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10 Jun 2020, 06:13
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IMO C
(A/Q) 2a^2+2b^2= 18x+y, y<18{a, b, x and y are non-negative integers}
Implies: y = remainder when (2a^2+2b^2) divided by 18 = 2 x [remainder (a^2+b^2) divided by 9]
Question: |y+3|=3 ? => y = 0 ? => [remainder (a^2+b^2) divided by 9]=0 ? (Since y=-6 not possible-cuz negative integer)
(1) The difference between a and b can be expressed as an even multiple of 3.
a-b = 3 x 2K (for some non-negative integer K)
a-b = {0,6,12,18...........}
Since a,b can take any values satisfying above eqn, e.q (3,3) or (7,1)
So, remainder (a^2+b^2) divided by 9 - Cannot be ascertain. (May be zero e.q (3,3) or may not be (7,1))
Not Sufficient
(2) b when divided by 3 is an integer
b= 3P (for some non-negative integer K)
But no idea about "a" -
Not Sufficient
Together:
a-b = 6K
a= 6K + 3P
a= 3(2K+P)
So, a^2 + b^2 = 9 [ (2K+P)^2 + P^2] = 9 x (Some integer)
Clearly, Rem. (a^2+b^2) divided by 9 = 0
Sufficient.
(A/Q) 2a^2+2b^2= 18x+y, y<18{a, b, x and y are non-negative integers}
Implies: y = remainder when (2a^2+2b^2) divided by 18 = 2 x [remainder (a^2+b^2) divided by 9]
Question: |y+3|=3 ? => y = 0 ? => [remainder (a^2+b^2) divided by 9]=0 ? (Since y=-6 not possible-cuz negative integer)
(1) The difference between a and b can be expressed as an even multiple of 3.
a-b = 3 x 2K (for some non-negative integer K)
a-b = {0,6,12,18...........}
Since a,b can take any values satisfying above eqn, e.q (3,3) or (7,1)
So, remainder (a^2+b^2) divided by 9 - Cannot be ascertain. (May be zero e.q (3,3) or may not be (7,1))
Not Sufficient
(2) b when divided by 3 is an integer
b= 3P (for some non-negative integer K)
But no idea about "a" -
Not Sufficient
Together:
a-b = 6K
a= 6K + 3P
a= 3(2K+P)
So, a^2 + b^2 = 9 [ (2K+P)^2 + P^2] = 9 x (Some integer)
Clearly, Rem. (a^2+b^2) divided by 9 = 0
Sufficient.
