NMTC 2015
a
,
b
,
c
,
d
,
e
are real numbers such that
a
+
4
b
+
9
c
+
1
6
d
+
2
5
e
=
1
4
a
+
9
b
+
1
6
c
+
2
5
d
+
3
6
e
=
8
9
a
+
1
6
b
+
2
5
c
+
3
6
d
+
4
9
e
=
2
3
.
Then the value of
a
+
b
+
c
+
d
+
e
is
Source: NMTC 2015 Inter Level.
The answer is 4.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Hi sir! @Otto Bretscher Is there any basis on your formula? can we always rely on that shortcut? thanks!
Log in to reply
Yes, there is a simple algebraic basis to this: If x , y , z are consecutive squares (such as 4,9,16 or 25,36,49), then x − 2 y + z = 2 .
@Vincent Jason Torres asked a great question. Comrade Otto, can you explain how you figure out what to add and what to subtract? Is it possible to generalize this?
Log in to reply
i did same sir. i did
α
n
2
+
β
(
n
+
1
)
2
+
γ
(
n
+
2
)
2
=
1
(
α
+
β
+
γ
)
n
2
+
(
2
β
+
4
γ
)
n
+
(
β
+
4
γ
)
=
1
⎩
⎪
⎨
⎪
⎧
α
+
β
+
γ
=
0
2
β
+
4
γ
=
0
β
+
4
γ
=
1
α
=
γ
=
.
5
,
β
=
−
1
Log in to reply
@Aareyan Manzoor – How did you get your very first equation?
Log in to reply
@Pi Han Goh – sir, the pattern is (for n =1) ⎩ ⎪ ⎨ ⎪ ⎧ n 2 a + ( n + 1 ) 2 b + . . ( n + 4 ) 2 e = 1 − − ( i ) ( n + 1 ) 2 a + ( n + 2 ) 2 b + . . ( n + 5 ) 2 e = 8 − − ( i i ) ( n + 2 ) 2 a + ( n + 3 ) 2 b + . . ( n + 6 ) 2 e = 2 3 − − ( i i i ) we want to have the coefficients=1. so, we need α ( i ) + β ( i i ) + γ ( i i i ) = ( α n 2 + β ( n + 1 ) 2 + γ ( n + 2 ) 2 ) a + ( α ( n + 1 ) 2 + β ( n + 2 ) 2 + γ ( n + 3 ) 2 ) b + . . . . ( α ( n + 4 ) 2 + β ( n + 5 ) 2 + γ ( n + 6 ) 2 ) e = a + b + c + d + e α k 2 + β ( k + 1 ) 2 + γ ( k + 2 ) 2 = 1 for k =n,n+1...,n+4
Log in to reply
@Aareyan Manzoor – Oh dang. I thought the solution poster have some magical technique. I see that it's just simple algebra.
Thank you very much! :D:D:D:D:D
Since 3 equations depend on 5 variables, values of any 2 variables can be assumed. I've assumed that d = e = 0 .
Therefore we get 3 equations of 3 variables:
-
a + 4 b + 9 c = 1
-
4 a + 9 b + 1 6 c = 1
-
9 a + 1 6 b + 2 5 c = 1
Therefore, on solving these 3 equations, a + b + c + d + e = 4
Taking (half of first equation) - (second equation) + (half of last equation) gives a + b + c + d + e = 2 1 − 8 + 2 2 3 = 4 .