Pre-RMO 2014/5

If real numbers a,b,c,d,ea, b, c, d, e satisfy

a+1=b+2=c+3=d+4=e+5=a+b+c+d+e+3,a + 1 = b + 2 = c + 3 = d + 4 = e + 5 = a + b + c + d + e + 3,

what is the value of a2+b2+c2+d2+e2a^2 + b^2 + c^2 + d^2 + e^2?


This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
4 years, 8 months ago

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Solving the equations a=2,b=1,c=o,d=-1,e=-2.Thus the answer is 10.

Vivek Rao - 4 years, 8 months ago

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But their square is 0

Sarita Goyal - 11 months, 1 week ago

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let this all be equal to k a=k-1 ,b=k-2 ,c=k-3, d=k-4 ,e=k-5 now substituting these values in last equation we get a,b,c,d,e

Vaibhav Sharma - 4 years, 7 months ago

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I did this sum by solving each part individually i.e I obtained every part in terms of d d and then I solved the equation to get the answer as 10.

mihir Chakravarti - 4 years, 6 months ago

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10

Sahil Nare - 4 years, 1 month ago

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10\color{red}{\boxed{10}}

Akshat Sharda - 3 years, 10 months ago

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10

gaurav singh - 3 years, 10 months ago

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10 😊

Swapnil Das - 3 years, 10 months ago

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let this all be equal to k a=k-1 ,b=k-2 ,c=k-3, d=k-4 ,e=k-5 now substituting these values in last equation we get a,b,c,d,e

nikhil jaiswal - 4 years, 8 months ago

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10

Soumya Ananthula - 4 years, 8 months ago

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10

Pooja Deshmukh - 4 years, 8 months ago

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the answer is 10..

Andy Leonardo - 4 years, 8 months ago

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Equate everything to k and substitute accordingly.The answer is 10.

Rohit Nair - 4 years, 8 months ago

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10

Aayush Solanki - 4 years, 8 months ago

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10

Zahra Y - 4 years, 8 months ago

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