38th EMC Blitz
Content
38th EMC Blitz Question 1(Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
3
Find the sum of all integer x so that xx2+8
−4 is an integer.
[Ans: 14]
Solution:
2
(x+2)(x −2x+4)
x(x−2)+4
x −2x+4
4
xx2+8
/ 2, 2)
−4 = (x−2)(x+2) = x−2 =
x−2 = x + x−2 (x =−
3
2
4 is an integer if 4 is also an integer,
x + x−2
x−2
x − 2 = {− 4,− 2,− 1, 1, 2, 4}
x = {− 2, 0, 1, 3, 4, 6} are the possible values
so the sum of all integer x is 0 + 1 + 3 + 4 + 6 = 14
38th EMC Blitz Question 2 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
∞
Evaluate the sum 1 + 2*12 + 3 122 + 4 123 + ... = ∑ n
*
1
2n−1
*
n=1
*
?
[Ans: ln(4) ]
Solution:
2
3
2
Let f (x) = 1 + 2x + 2x2 + 2x3 + ... = 1−1x = ( 2−x1 ) = 2−x
2
2
Integrate f (x) ,
2 dx
∫ f (x)dx = x + 2x*2 + 3*x2 + 4*x2 ... = ∫ 2−x
2
3
4
2
3
If f (x) evaluate from x = 0 to x = 1
1
2
3
4
1
2 dx = − 2ln(2 − x) x=1
∫ f (x)dx = x + 2x*2 + 3*x2 + 4*x2 ... = 1 + 21*2 + 3*12 + 4*12 ... = ∫ 2−x
x=0
0
2
3
4
2
3
= − 2ln(1) + 2ln(2) = ln(4)
2
3
0
38th EMC Blitz Question 3 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Calculate the area of the region bounded by the curve y = x2014 and x = y2014 and lying above the
xaxis?
[Ans. 2013/2015]
Solution:
The two curves intersect at (0,0) and (1,1),
1
1
A = ∫ x 2014 − x2014dx =
0
1
2015
x2014 − x2015
2015
2015
2014
0
[
]
1
2013
= 2014
2015 − 2015 = 2015
38th EMC Blitz Question 4 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Given the three roots 2012,− 2013, 2014, of f (x) = x4 − ax2 + bx − c , find the value of a + b + c?
[Ans. 1 − 20122(2013)(2015) ]
Solution:
Note that f (x) = (x − 2012)(x + 2013)(x − 2014)(x − x4) = x4 − ax2 + bx − c
also, 2012 + (− 2013) + 2014 + x4 = 0, x4 = − 2013
f (x) = (x − 2012)(x + 2013)(x − 2014)(x + 2013)
f (− 1) = 1 − a − b − c = (− 1 − 2012)(− 1 + 2013)(− 1 − 2014)(− 1 + 2013) = (− 2013)(2012)(− 2015)(2012)
1 − f (− 1) = a + b + c = 1 − 20122(2013)(2015)
38th EMC Blitz Question 5 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Urn A contains 2 pink and 2 violet balls and Urn B contains 3 pink and 2 violet balls. One ball is chosen at
random from Urn A and transferred to Urn B, and then a ball is chosen at random from Urn B. The ball
chosen from Urn B is observed to be pink. Find the probability that the ball transferred from Urn A to
Urn B was pink.
[Ans. 4/7 ]
Solution:
Let X denote the event that the ball transferred from Urn A to Urn B was pink.
Let Y denote the event that the ball chosen from Urn B is observed to be pink.
We need to find P(X|Y).
Observe that P(X) = ½ and P(X’) = ½
If the ball transferred is pink, P(Y|X) = 4/6 = ⅔
If the ball transferred is violet, P(Y|X’) = 3/6 = ½
P (X ⋂ Y ) = P (Y |X)P (X) = (½)(⅔) = ⅓
P (X ′ ⋂ Y ) = P (Y |X ′)P (X ′) =(½)(½) = ¼
P (Y ) = P (X ⋂ Y ) + P (X ′ ⋂ Y ) = ⅓ +¼ = 7/12
)
1/3
4/12
P (X|Y ) = P(X⋂Y
P(Y ) = 7/12 = 7/12 = 4/7
38th EMC Blitz Question 6 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
1
The number of injury claims per month is modeled by a random variable N with P [N = n] = (n+1)(n+2)
where n ≥ 0 . Determine the probability of at least one claim during a particular month, given that there
have been at most four claims during that month.
[Ans. 25 ]
Solution:
P [N ≥ 1|N ≤ 4] = P[1≤N≤4]
P [N≤4]
P [1 ≤ N ≤ 4] = P [N = 1] + P [N = 2] + P [N = 3] + P [N = 4] = 2*13 + 3*14 + 4*15 + 5*16 = 13
P [N ≤ 4] = P [N = 0] + P [1 ≤ N ≤ 4] = 1*12 + 13 = 56
P [N ≥ 1|N ≤ 4] = (1/3)
(5/6) =
(2/6)
(5/6)
= 25
38th EMC Blitz Question 7 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Given T 1 = 0, T 2 = 1, T 3 = 1 and satisfy the equation T n = T n−1 + T n−2 + T n−3 where n ≥ 4.
n
T
Calculate lim ∑ 2014kk .
n→∞ k=1
2014
[Ans.20143−2014
]
2
−2014−1
Solution:
n
T
Let S = lim ∑ xkk
n→∞ k=1
T k = {0, 1, 1, 2, 4, 7, 13, 24, 44, ...}
S = x01 + x12 + x13 + x24 + x45 + x76 + 13
x7 + ...
24
S x = x11 + x12 + x23 + x44 + x75 + 13
x6 + x7 + ...
24
44
S x2 = 1 + 1x + x22 + x43 + x74 + 13
x5 + x6 + x7 + ...
S x2 S x S = 1 + x13 + x14 + x25 + x46 + x77 + ...
x (S x2 S x S) = x + x12 + x13 + x24 + x45 + x76 + ...
S( x3 − x2 − x ) = x + S
S( x3 − x2 − x − 1 ) = x
S = x3−x2x−x−1
Let x = 2014
2014
S = 20143−2014
2
−2014−1
38th EMC Blitz Question 8 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Let R(x) be the remainder when x2013
+ 2014 is divided by x2 − 1 . If
2015
R(x)−x
R(x)+x dx =
1
∫
cln(ab) , where
gcd(a,b) = 1 and c is a positive integer. What is the value of a b + c?
[Ans: 2014]
Solution:
P(x) = Q(x)D(x) + R(x)
D(x) = x2 − 1
R(x) = ax + b (since D(x) is a 2nd degree polynomial so 2 - 1 = 1st degree)
x2013 + 2014 = Q(x)(x2 − 1) + ax + b
if x = 1,
2015 = a + b
if x = -1
2013 = -a + b
2b = 4028, b = 2014
2a = 2, a = 1
R(x) = x + 2014
Now,
=
2015
R(x)−x
R(x)+x dx
1
∫
=
2015
x+2014−x dx
x+2014+x
1
∫
1007ln(x + 1007) x=2015
x=1
= 2014
2015
∫
1
1
2x+2014 dx
= 1007
2015
∫
1
1
x+1007 dx
1511
= 1007(ln(2015 + 1007) − ln(1 + 1007)) = 1007ln(3022
1008) =1007ln( 504 )
since gcd(1511,504) = 1, then a − b = 1511 − 504 = 1007
Then (a - b) + c = 1007 + 1007 = 2014
38th EMC Blitz Question 9 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Evaluate 1 +
√
√1 + 3√1 + 4√1 + ...
1+2
[Ans. 4]
Solution:
√(x + 2)
Note that x√1 + (x + 1)(x + 3) = x√1 + x2 + 4x + 3 = x√x2 + 4x + 4 = x
for x ≥ 0
Let f (x) = x(x + 2) = x√1 + (x + 1)(x + 3)
f (x) = x√1 + (x + 1)(x + 3)
f (x) = x√1 + f (x + 1)
√1 + (x + 1)√1 + (x + 2)(x + 4)
f (x) = x√1 + (x + 1)√1 + f (x + 2)
f (x) = x 1 + (x + 1) 1 + (x + 2)√1 + (x + 3)(x + 5)
√
√
f (x) = x 1 + (x + 1) 1 + (x + 2)√1 + ... = x(x + 2)
√
√
f (x) = x
Let x = 1
f (1) =
√
√
1 + (1 + 1)
1 + f (1) = 1 +
√
1 + (1 + 2)
√1 + (1 + 3)√1 + ... = 1(1 + 2) = 3
√1 + 3√1 + 4√1 + ... = 1 + 3 = 4
1+2
2
= x(x + 2)
38th EMC Blitz Question 10 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Compute the volume of Tetrahedron ABCD given the vertices A(2011, 1, 4), B(3017, 1000, 3),
C(3, 10, 2021) and D(2016, 1, 1).
[Ans. 0 units3 ]
Solution:
The 4 vertices lie on the same plane x y = z + 2014, thus the 4 vertices are coplanar and the volume is 0.
March 29, 2014 (8:00 PM +8 GMT)
3
Find the sum of all integer x so that xx2+8
−4 is an integer.
[Ans: 14]
Solution:
2
(x+2)(x −2x+4)
x(x−2)+4
x −2x+4
4
xx2+8
/ 2, 2)
−4 = (x−2)(x+2) = x−2 =
x−2 = x + x−2 (x =−
3
2
4 is an integer if 4 is also an integer,
x + x−2
x−2
x − 2 = {− 4,− 2,− 1, 1, 2, 4}
x = {− 2, 0, 1, 3, 4, 6} are the possible values
so the sum of all integer x is 0 + 1 + 3 + 4 + 6 = 14
38th EMC Blitz Question 2 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
∞
Evaluate the sum 1 + 2*12 + 3 122 + 4 123 + ... = ∑ n
*
1
2n−1
*
n=1
*
?
[Ans: ln(4) ]
Solution:
2
3
2
Let f (x) = 1 + 2x + 2x2 + 2x3 + ... = 1−1x = ( 2−x1 ) = 2−x
2
2
Integrate f (x) ,
2 dx
∫ f (x)dx = x + 2x*2 + 3*x2 + 4*x2 ... = ∫ 2−x
2
3
4
2
3
If f (x) evaluate from x = 0 to x = 1
1
2
3
4
1
2 dx = − 2ln(2 − x) x=1
∫ f (x)dx = x + 2x*2 + 3*x2 + 4*x2 ... = 1 + 21*2 + 3*12 + 4*12 ... = ∫ 2−x
x=0
0
2
3
4
2
3
= − 2ln(1) + 2ln(2) = ln(4)
2
3
0
38th EMC Blitz Question 3 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Calculate the area of the region bounded by the curve y = x2014 and x = y2014 and lying above the
xaxis?
[Ans. 2013/2015]
Solution:
The two curves intersect at (0,0) and (1,1),
1
1
A = ∫ x 2014 − x2014dx =
0
1
2015
x2014 − x2015
2015
2015
2014
0
[
]
1
2013
= 2014
2015 − 2015 = 2015
38th EMC Blitz Question 4 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Given the three roots 2012,− 2013, 2014, of f (x) = x4 − ax2 + bx − c , find the value of a + b + c?
[Ans. 1 − 20122(2013)(2015) ]
Solution:
Note that f (x) = (x − 2012)(x + 2013)(x − 2014)(x − x4) = x4 − ax2 + bx − c
also, 2012 + (− 2013) + 2014 + x4 = 0, x4 = − 2013
f (x) = (x − 2012)(x + 2013)(x − 2014)(x + 2013)
f (− 1) = 1 − a − b − c = (− 1 − 2012)(− 1 + 2013)(− 1 − 2014)(− 1 + 2013) = (− 2013)(2012)(− 2015)(2012)
1 − f (− 1) = a + b + c = 1 − 20122(2013)(2015)
38th EMC Blitz Question 5 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Urn A contains 2 pink and 2 violet balls and Urn B contains 3 pink and 2 violet balls. One ball is chosen at
random from Urn A and transferred to Urn B, and then a ball is chosen at random from Urn B. The ball
chosen from Urn B is observed to be pink. Find the probability that the ball transferred from Urn A to
Urn B was pink.
[Ans. 4/7 ]
Solution:
Let X denote the event that the ball transferred from Urn A to Urn B was pink.
Let Y denote the event that the ball chosen from Urn B is observed to be pink.
We need to find P(X|Y).
Observe that P(X) = ½ and P(X’) = ½
If the ball transferred is pink, P(Y|X) = 4/6 = ⅔
If the ball transferred is violet, P(Y|X’) = 3/6 = ½
P (X ⋂ Y ) = P (Y |X)P (X) = (½)(⅔) = ⅓
P (X ′ ⋂ Y ) = P (Y |X ′)P (X ′) =(½)(½) = ¼
P (Y ) = P (X ⋂ Y ) + P (X ′ ⋂ Y ) = ⅓ +¼ = 7/12
)
1/3
4/12
P (X|Y ) = P(X⋂Y
P(Y ) = 7/12 = 7/12 = 4/7
38th EMC Blitz Question 6 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
1
The number of injury claims per month is modeled by a random variable N with P [N = n] = (n+1)(n+2)
where n ≥ 0 . Determine the probability of at least one claim during a particular month, given that there
have been at most four claims during that month.
[Ans. 25 ]
Solution:
P [N ≥ 1|N ≤ 4] = P[1≤N≤4]
P [N≤4]
P [1 ≤ N ≤ 4] = P [N = 1] + P [N = 2] + P [N = 3] + P [N = 4] = 2*13 + 3*14 + 4*15 + 5*16 = 13
P [N ≤ 4] = P [N = 0] + P [1 ≤ N ≤ 4] = 1*12 + 13 = 56
P [N ≥ 1|N ≤ 4] = (1/3)
(5/6) =
(2/6)
(5/6)
= 25
38th EMC Blitz Question 7 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Given T 1 = 0, T 2 = 1, T 3 = 1 and satisfy the equation T n = T n−1 + T n−2 + T n−3 where n ≥ 4.
n
T
Calculate lim ∑ 2014kk .
n→∞ k=1
2014
[Ans.20143−2014
]
2
−2014−1
Solution:
n
T
Let S = lim ∑ xkk
n→∞ k=1
T k = {0, 1, 1, 2, 4, 7, 13, 24, 44, ...}
S = x01 + x12 + x13 + x24 + x45 + x76 + 13
x7 + ...
24
S x = x11 + x12 + x23 + x44 + x75 + 13
x6 + x7 + ...
24
44
S x2 = 1 + 1x + x22 + x43 + x74 + 13
x5 + x6 + x7 + ...
S x2 S x S = 1 + x13 + x14 + x25 + x46 + x77 + ...
x (S x2 S x S) = x + x12 + x13 + x24 + x45 + x76 + ...
S( x3 − x2 − x ) = x + S
S( x3 − x2 − x − 1 ) = x
S = x3−x2x−x−1
Let x = 2014
2014
S = 20143−2014
2
−2014−1
38th EMC Blitz Question 8 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Let R(x) be the remainder when x2013
+ 2014 is divided by x2 − 1 . If
2015
R(x)−x
R(x)+x dx =
1
∫
cln(ab) , where
gcd(a,b) = 1 and c is a positive integer. What is the value of a b + c?
[Ans: 2014]
Solution:
P(x) = Q(x)D(x) + R(x)
D(x) = x2 − 1
R(x) = ax + b (since D(x) is a 2nd degree polynomial so 2 - 1 = 1st degree)
x2013 + 2014 = Q(x)(x2 − 1) + ax + b
if x = 1,
2015 = a + b
if x = -1
2013 = -a + b
2b = 4028, b = 2014
2a = 2, a = 1
R(x) = x + 2014
Now,
=
2015
R(x)−x
R(x)+x dx
1
∫
=
2015
x+2014−x dx
x+2014+x
1
∫
1007ln(x + 1007) x=2015
x=1
= 2014
2015
∫
1
1
2x+2014 dx
= 1007
2015
∫
1
1
x+1007 dx
1511
= 1007(ln(2015 + 1007) − ln(1 + 1007)) = 1007ln(3022
1008) =1007ln( 504 )
since gcd(1511,504) = 1, then a − b = 1511 − 504 = 1007
Then (a - b) + c = 1007 + 1007 = 2014
38th EMC Blitz Question 9 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Evaluate 1 +
√
√1 + 3√1 + 4√1 + ...
1+2
[Ans. 4]
Solution:
√(x + 2)
Note that x√1 + (x + 1)(x + 3) = x√1 + x2 + 4x + 3 = x√x2 + 4x + 4 = x
for x ≥ 0
Let f (x) = x(x + 2) = x√1 + (x + 1)(x + 3)
f (x) = x√1 + (x + 1)(x + 3)
f (x) = x√1 + f (x + 1)
√1 + (x + 1)√1 + (x + 2)(x + 4)
f (x) = x√1 + (x + 1)√1 + f (x + 2)
f (x) = x 1 + (x + 1) 1 + (x + 2)√1 + (x + 3)(x + 5)
√
√
f (x) = x 1 + (x + 1) 1 + (x + 2)√1 + ... = x(x + 2)
√
√
f (x) = x
Let x = 1
f (1) =
√
√
1 + (1 + 1)
1 + f (1) = 1 +
√
1 + (1 + 2)
√1 + (1 + 3)√1 + ... = 1(1 + 2) = 3
√1 + 3√1 + 4√1 + ... = 1 + 3 = 4
1+2
2
= x(x + 2)
38th EMC Blitz Question 10 (Hosted by Lawrence)
March 29, 2014 (8:00 PM +8 GMT)
Compute the volume of Tetrahedron ABCD given the vertices A(2011, 1, 4), B(3017, 1000, 3),
C(3, 10, 2021) and D(2016, 1, 1).
[Ans. 0 units3 ]
Solution:
The 4 vertices lie on the same plane x y = z + 2014, thus the 4 vertices are coplanar and the volume is 0.
