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08 Feb 2021, 07:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
79% (01:20) correct
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If a, b, c, and d are consecutive odd numbers, then \(a^2 + b^2 + c^2 + d^2\) is always divisible by:
A. 3
B. 4
C. 5
D. 6
E. 7
A. 3
B. 4
C. 5
D. 6
E. 7
08 Feb 2021, 20:32
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Bunuel
All 4 terms will be odd, so answer has to be even. Only 4 and 6 possible.
Now \(1^2+3^2+5^2+7^2\) and \(3^2+5^2+7^2+9^2\) will have different properties when divided by 3, as 1^2, a number not divisible by 3, is replaced by 9^2, a number divisible by 3.
So 4 is the answer.
We can also check by substituting the odd number as 2n-3, 2n-1, 2n+1 and 2n+3.
\((2n-3)^2+(2n-1)^2+(2n+1)^2+(2n+3)^2\)
\(4n^2-12n+9+4n^2-4n+1+4n^2+4n+1+4n^2+12n+9=16n^2+20=4(4n^2+5)\)
So divisible by 4.
B
08 Feb 2021, 20:51
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Let the numbers be \((x+1),(x+3),(x+5),(x+7)\)
\((x+1)^2+(x+3)^2+(x+5)^2+(x+7)^2 = (x^2+2x+1)+(x^2+6x+9)+(x^2+10x+25)+(x^2+14x+49)=4x^2+32x+84=4(x^2+8x+21)\)
Therefore, it is always divisible by \(4\)
Ans B
Quick Approach; \(1^2+3^2+5^2+7^2=1+9+25+49=84=4*21\)
\((x+1)^2+(x+3)^2+(x+5)^2+(x+7)^2 = (x^2+2x+1)+(x^2+6x+9)+(x^2+10x+25)+(x^2+14x+49)=4x^2+32x+84=4(x^2+8x+21)\)
Therefore, it is always divisible by \(4\)
Ans B
Quick Approach; \(1^2+3^2+5^2+7^2=1+9+25+49=84=4*21\)
