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If a and b are positive integers, is a^2 + b^2 divisible by 5? 30 Sep 2017, 01:43
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C
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67% (01:38) correct 33% (01:34) wrong based on 208 sessions

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If \(a\) and \(b\) are positive integers, is \(a^2 + b^2\) divisible by 5?


(1) \(2ab\) is divisible by 5

(2) \(a - b\) is divisible by 5


M19-11

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C
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If a and b are positive integers, is a^2 + b^2 divisible by 5? 30 Sep 2017, 02:46
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Bunuel
If \(a\) and \(b\) are positive integers, is \(a^2 + b^2\) divisible by 5?


(1) \(2ab\) is divisible by 5

(2) \(a - b\) is divisible by 5


M19-11
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Statement (1) says tat 2ab is divisible by 5, hence option is insufficient.
Statement (2) Doesnt say anything about the value of a & b hence insufficient

taking both options together we get : (a-d)^2 = 25 => a^2 + b^2 - 2ab = 25

=> a^2 + b^2 = 25 + 2ab

since both 25 and 2ab is divisible by 5 we can conclude that a^2 + b^2 is divisible by 5
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If a and b are positive integers, is a^2 + b^2 divisible by 5? 30 Sep 2017, 02:47
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Bunuel
If \(a\) and \(b\) are positive integers, is \(a^2 + b^2\) divisible by 5?


(1) \(2ab\) is divisible by 5

(2) \(a - b\) is divisible by 5


M19-11
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Statement 1: if \(a=5\) and \(b=1\), then \(2ab\) is divisible by \(5\) but \(a^2+b^2\) is not, where as if \(a=b=5\), then both \(2ab\) and \(a^2+b^2\) is divisible by \(5\). Hence Insufficient

Statement 2: if \(a=b=5\), then both \(a-b\) & \(a^2+b^2\) is divisible by \(5\), but if \(a=6\) and \(b=1\), then \(a-b\) is divisible by \(5\) but \(a^2+b^2\) is not. Hence Insufficient

Combining 1 & 2, we can write \(a^2+b^2=(a-b)^2+2ab\), Now \(a-b\) is divisible by \(5\) hence its square will also be divisible by \(5\) and \(2ab\) is divisible by \(5\). So we can say that \(a^2+b^2\) is divisible by \(5\). Hence Sufficient

Option C
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If a and b are positive integers, is a^2 + b^2 divisible by 5? 30 Sep 2017, 05:25
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Bunuel
If \(a\) and \(b\) are positive integers, is \(a^2 + b^2\) divisible by 5?


(1) \(2ab\) is divisible by 5

(2) \(a - b\) is divisible by 5


M19-11
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Yes no type Qs, a
Given info- a and b are positive integer then a2+b2/5 =integer??

St-1 2ab is divisible by 5, then a is multiple of 5 or b is multiple of 5 or both a & b is multiple of 5.
If a& b both multiple of 5 then sufficient if a only or b only is multiple then insufficient.

St-2 a-b is multiple of 5, a and b both multiple of 5 then sufficient also a and b can not be multiple of 5 eg. 8-3 is divisible by 5. insufficient.

Combine--If either a or b is multiple of 5 then other one has to be multiple of 5 to satisfy a-b is multiple of 5. then sufficient.
if a is 10 and b is 2 then a-b can not be divisible by 5. for both statement true and a and b has to be multiple of 5.

Answer is C
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If a and b are positive integers, is a^2 + b^2 divisible by 5? 24 May 2023, 22:48
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Bunuel
If \(a\) and \(b\) are positive integers, is \(a^2 + b^2\) divisible by 5?
(1) \(2ab\) is divisible by 5
(2) \(a - b\) is divisible by 5
Show more
Solution:
Pre Analysis:
  • \(a\) and \(b\) are positive integers
  • We are asked if \(a^2 + b^2\) divisible by 5 or not

Statement 1: \(2ab\) is divisible by 5
  • According to this statement, \(2ab=5k\) where k is a positive integer
  • However, this doesn't tell us if \(a^2+b^2\) is divisible by 5 or not
  • Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: \(a - b\) is divisible by 5
  • According to this statement, \(a-b=5q\) where q is a positive integer
  • Squaring both sides we have \((a-b)^2=(5q)^2\)
    \(⇒a^2+b^2-2ab=25q^2\)
    \(⇒a^2+b^2=25q^2+2ab\)
  • However, this doesn't tell us if \(a^2+b^2\) is divisible by 5 or not without knowing if \(2ab\) is divisible by 5 or not
  • Thus, statement 2 alone is not sufficient

Combining:
  • From statement 1, \(2ab=5k\)
  • From statement 2, \(a^2+b^2=25q^2+2ab\)
  • So, \(a^2+b^2=25q^2+2ab=25q^2+5k=5(5q^2+k)\)
  • Which means \(a^2+b^2\) is divisible by 5

Hence the right answer is Option C
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If a and b are positive integers, is a^2 + b^2 divisible by 5? 01 Oct 2023, 10:56
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If \(a\) and \(b\) are positive integers, is \(a^2 + b^2\) divisible by 5?

(1) \(2ab\) is divisible by 5.

This implies that either \(a\), \(b\), or both are divisible by 5. If \(a = b = 5\), the answer is YES. However, \(a = 5\) and \(b = 1\), the answer is NO. Not sufficient.

(2) \(a-b\) is divisible by 5.

This implies that either both \(a\) and \(b\) are divisible by 5 or neither are. If \(a = b = 5\), the answer is YES. However, \(a = b = 1\), the answer is NO. Not sufficient.

(1)+(2) From statement (2), \(a - b\) is divisible by 5, so \((a-b)^2=(a^2+b^2)-2ab\) is also divisible by 5. Since from statement (1) \(2ab\) is divisible by 5, then \(a^2 + b^2\) must also be divisible by 5 for their difference to be divisible by 5. Sufficient.

Alternatively, from statement (2), if neither \(a\) nor \(b\) is divisible by 5, it would contradict statement (1). Thus, the only remaining scenario from statement (2) is that both \(a\) and \(b\) are divisible by 5. Consequently, \(a^2 + b^2\) is also divisible by 5. Sufficient.


Answer: C
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