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Pre-RMO 2014/5

If real numbers \(a, b, c, d, e\) satisfy

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

what is the value of \(a^2 + b^2 + c^2 + d^2 + e^2\)?


This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
2 years, 9 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 · 2 years, 9 months ago

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10 😊 Swapnil Das · 1 year, 11 months ago

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10 Gaurav Singh · 1 year, 11 months ago

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\(\color{red}{\boxed{10}}\) Akshat Sharda · 1 year, 11 months ago

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10 Sahil Nare · 2 years, 2 months ago

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I did this sum by solving each part individually i.e I obtained every part in terms of \( d \) and then I solved the equation to get the answer as 10. Mihir Chakravarti · 2 years, 7 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 Vaibhav Sharma · 2 years, 9 months ago

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the answer is 10.. Andy Leonardo · 2 years, 9 months ago

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10 Pooja Deshmukh · 2 years, 9 months ago

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10 Soumya Ananthula · 2 years, 9 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 · 2 years, 9 months ago

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Equate everything to k and substitute accordingly.The answer is 10. Rohit Nair · 2 years, 9 months ago

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10 Zahra Y · 2 years, 9 months ago

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10 Aayush Solanki · 2 years, 9 months ago

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