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

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestSolving the equations a=2,b=1,c=o,d=-1,e=-2.Thus the answer is 10. – Vivek Rao · 2 years, 5 months ago

Log in to reply

10 😊 – Swapnil Das · 1 year, 7 months ago

Log in to reply

10 – Gaurav Singh · 1 year, 7 months ago

Log in to reply

\(\color{red}{\boxed{10}}\) – Akshat Sharda · 1 year, 7 months ago

Log in to reply

10 – Sahil Nare · 1 year, 11 months ago

Log in to reply

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, 3 months ago

Log in to reply

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, 5 months ago

Log in to reply

the answer is 10.. – Andy Leonardo · 2 years, 5 months ago

Log in to reply

10 – Pooja Deshmukh · 2 years, 5 months ago

Log in to reply

10 – Soumya Ananthula · 2 years, 5 months ago

Log in to reply

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, 5 months ago

Log in to reply

Equate everything to k and substitute accordingly.The answer is 10. – Rohit Nair · 2 years, 5 months ago

Log in to reply

10 – Zahra Y · 2 years, 5 months ago

Log in to reply

10 – Aayush Solanki · 2 years, 5 months ago

Log in to reply