Individual Questions
(1)
Let f(n) be the sum of a positive integer n and its digits. Find all numbers such that f(n) =
2010.
The number n must be 1xyz or 2xyz, where x, y, z are its hundreds, tens, and units digits,
respectively. Since x + y + z  27 and 2010 – 27 = 1983, we must start with a least 198z.
1980 + z + 1 + 9 + 8 + z = 1998 + 2z = 2010  2z = 12  z = 6 and f(1986) = 2010
Next, try 199z. 1990 + z + 1 + 9 + 9 + z is odd, no such numbers.
Next, try 200z. 2000 + z + 2 + z = 2010  2z = 8  z = 4  f(2004) = 2010
So, 1986 and 2004 are the only solutions.
(2)
 x2  y 2
For which k does the system 
2
2
( x  k )  y
0
has exactly two real solutions?
2
The first equation implies y2 = x2. Substituting this in the second equation, we have
2x2 – 2kx + k2 – 2 = 0
To have two real solutions  only one x solution because there are two y’s for each x > 0
 (2k)2 – 4(2)(k2 – 2) = 0
 4k2 – 8k2 + 16 = 0
 k2 = 4
 k= 2 or –2
(3)
How many pairs of positive integers have greatest common divisor 3! and least common
multiple 18!?
3! = 2∙3, 18! = 2∙3∙4∙5∙6∙7∙8∙9∙10∙11∙12∙13∙14∙15∙16∙17∙18 = 216∙38∙53∙72∙11∙13∙17
Both numbers have 2∙3 as a factor.
Both numbers can’t simultaneously have a higher power of 2 or 3 as factor.
One number must have 216 as factor and the other one only has 2 as factor.
One number must have 38 as factor and the other one only has 3 as factor.
One number must have 53 as factor and the other one doesn’t have 5 as factor.
One number must have 72 as factor and the other one doesn’t have 7 as factor.
One number must have 11 as factor and the other one doesn’t have 11 as factor.
One number must have 13 as factor and the other one doesn’t have 13 as factor.
One number must have 17 as factor and the other one doesn’t have 17 as factor.
Thought there are 27 combinations, but one pair consists of two combinations. So there
are 26 pairs.
(4)
Suppose you have a piece of paper which you cut into either four or sixteen pieces. After
that you cut again some of the pieces into either four or sixteen smaller pieces. Suppose
you have nothing else to do, so you keep repeating this procedure cutting some of the
pieces into either four or sixteen smaller pieces. Can you end up with 2010 pieces at some
stage of your cutting process?
Every time we increase the number of pieces by 3k + 15l, where k is the number of pieces
that we cut into four, and l is the number of pieces we cut into sixteen.
Since we start with one piece of paper, in order to have 2010 pieces we must have
2010 = 1 + 3k + 15l for some natural numbers k and l.
Thus 2009 = 3k + 15l, which is impossible since 3 does not divide 2009.
(5)
A group of 150 high school students visited Missouri State University for Computer
Science, Engineering, and Math Day. The students could participate in at most two of the
three Science, Engineering, and Math workshops but must attend at least one. Suppose
100 students participated in Computer Science, 90 participated in Engineering, and 80
participated in Math. If 87 participated in both Computer Science and Engineering or in
both Computer Science and Math. How many students participated in both Engineering
and Math?
Let C, E, and M denote the sets of students participating, correspondingly, in the
Computer Science, Engineering, and Math workshops. We then have
C E M  C  E  M  C E  C M  E M  C E M
 150 = 100 + 90 + 80 – 87 – E  M + 0
 E  M = 33
(6)
Find the smallest positive integer N such that
1
1
N is a perfect cube and N is a perfect
3
7
seventh power.
1
1
N  a 3 , N  b 7 for some positive integers a, and b  N = 3a3 = 7b7
3
7
 a is divisible by 7 and b is divisible by 3
 The smallest possible N should only have factors 3 and 7.
 N = 3(3h7k)3 = 7(3r7t)7 for some h, k, r, and t
 N = 33h+173k = 37r77t+1 for some h, k, r, and t
 3h + 1 = 7r and 3k = 7t + 1, and the smallest possible solutions h = 2, r = 1, k = 5, t = 1
 N = 37715
(7)
How many ways are there to assign the numbers 1, 3, 5, 7, and 9 to c, o, k, i, and e so that
the numerical equivalent to ‘cookie’ is divisible by 3?
A number is divisible by 3 if its digit sum is divisible by 3, so c + 2o + k + i + e is
divisible by 3. Since 25 + o = 1 + 3 + 5 + 7 + 9 + o = c + o + k + i + e + o must be
divisible by 3, o can only be 5. There are no other restrictions on c, k, i, and e, so there are
4! = 24 possibilities.
(8)
If x + y + z = 5, xy + xz + yz = 6, and xyz = 4, find x3 + y3 + z3.
( x  y  z )3  x 3  y 3  z 3  3( x 2 y  xy 2  x 2 z  xz 2  y 2 z  yz 2 )  6 xyz
 x 3  y 3  z 3  3[ xy ( x  y )  xz ( x  z )  yz ( y  z )]  6 xyz
 x 3  y 3  z 3  3[ xy (5  z )  xz (5  y )  yz (5  x)]  6 xyz
 x 3  y 3  z 3  3[5( xy  xz  yz )  ( xyz  xyz  xyz )]  6 xyz
x 3  y 3  z 3  ( x  y  z )3  3[5( xy  xz  yz )  ( xyz  xyz  xyz )]  6 xyz
  125  3(5  6  12)  6  4
 125  54  24  47
(9)
Suppose x3 – 6x + k and x3 – 10x + 2k have a common root, What must k be?
If r is a common root, then r3 – 6r + k = 0 and r3 – 10r + 2k = 0
 k = r3 – 6r and 2k = r3 – 10r
 2r3 – 12r = r3 – 10r OR r3 – 2r = 0
 r = 0, 2 ,  2
 k = 0, –4 2 , 4 2
(10) A polynomial P(x) of degree three is known to have P(1) = 1, P(2) = 1, P(3) = 2, and P(4)
= 2. Compute P(5)
( x  2)( x  3)( x  4)
( x  1)( x  3)( x  4)
 1
(1  2)(1  3)(1  4)
(2  1)(2  3)(2  4)
( x  1)( x  2)( x  4)
( x  1)( x  2)( x  3)
2 
 2
(3  1)(3  2)(3  4)
(4  1)(4  2)(4  3)
P( x)  1 
(5  2)(5  3)(5  4)
(5  1)(5  3)(5  4)
 1
(1  2)(1  3)(1  4)
(2  1)(2  3)(2  4)
(5  1)(5  2)(5  4)
(5  1)(5  2)(5  3)
 2 
 2
(3  1)(3  2)(3  4)
(4  1)(4  2)(4  3)
P(5)  1
(3)(2)(1)
(4)(2)(1)
(4)(3)(1)
(4)(3)(2)

 2
 2
(1)(2)(3) (1)(1)(2)
(2)(1)(1)
(3)(2)(1)
 1  4  2  6  2  4  1
