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OHMIO Ciphering 2007 Questions and Solutions
Practice:
What positive integer is equal to: (15325241)(15325241)-(15325233)(15325249)
Practice
Answer: 64
Practice:
Solution: Let x = 15325241, then x2   x  8 x  8  x2  x2  64  64
1. Write 101  100 105  104  103  102  101  100  as an integer in simplest form.
1. Answer: 999,999
1. Solution:
Since  x  1  x5  x 4  x3  x 2  x  1  x 6  1
when x  1 the given product = 106  1  999,999
2. Each valve A, B, and C when open, releases water into a tank at its own constant rate. With
all three valves open, the tank fills in 1 hour, with only valves A and C open it takes 1.5 hours,
and with only valves B and C open it takes 2 hours. The number of hours required with only
valves A and B open is what?
2. Answer: 1.2 hours
2. Solution:
a  rate A fills
b  rate B fills
c  rate C fills
1 a+b+c   1
a b  c 1
3
a  c  1
2
ac 
2
3
1
bc 
2
2 b  c   1
1
2
1
b
3
a
ab 
5
6
6
a  b  1
5
A and B fill in
6
or 1.2 hours
5
2x 4 y  1
3. If
4x 2 y 
x y ?
then
3. Answer:
1
2
1
3
3. Solution:
2 x 22 y  1
2x  22 y
x  2 y
22 x 2 y  21
22x  2 y 1
2x   y  1
 4 y   y 1
y
1
3
x
2
3
x y 
1
3
OHMIO Ciphering 2007 Questions and Solutions
2
4. The graph f  x   x 3 is plotted on the domain  0, a  where a  0 . The area between f  x 
and the x -axis is revolved 360  about the y -axis such that a solid with volume
3
is
64
generated. Find the exact value of a in simplified form..
4. Answer:
2
4
4. Solution: Using the shell method:
2 
a
0
8 a
3 3
x x dx 
x
4
2
3
0
3 3 3
a 
,
4
64
8

3
8
a 3  24 , a  2 2 
2
4
5. Find the set of four consecutive positive integers such that the smallest is a multiple of 5, the
second is a multiple of 7, the third is a multiple of 9, and the largest is a multiple of 11.
5. Answer: 1735, 1736, 1737, 1738
5. Solution:
Let M  5  7  9  11  3465
These numbers have the property that 5 divides P, 7 divides Q,
Let P  M  5
9 divides R, and 11 divides S. Since M is odd, these numbers
Q  M 7
are consecutive even numbers. However dividing by 2 gives the
R  M 9
required set:
1735, 1736, 1737, 1738
S  M  11
6. Find the sum of all solutions on the interval 0, 2  of the equation:
cos  2 x   cos  x 
6. Answer: 4
6. Solution:
cos  2 x   2 cos 2 x  1  cos x
2 cos 2 x  cos x  1  0
 2 cos x  1 cos x  1  0
cos x  
1
2
cos x  1
2 4
,
, 2
3
3
2 4
sum  0 

 2  4
3
3
x  0,
OHMIO Ciphering 2007 Questions and Solutions
7. The sum of all but one of the interior angles of a convex polygon equals 2570  . The
remaining angle contains how many degrees?
7. Answer: 130 
7. Solution:
 n  2 180  2570  x
where 0  x  180. If n  17, then  n  2 180  15 180  =2700 and
x  2700  2570  130. If n  17, then x  0, and if n  17, then x  180.
8. Shelby is running late one day and has no time to pick out her socks, which are jammed into
her drawer, not in pairs. She knows, if matched, she has 2 pairs of red socks, 4 pairs of blue
socks, 3 pairs of gray socks, 9 pairs of white socks, and 1 pair of orange socks. Then, she has 1
purple sock which has lost its mate. She knows the first sock she put on is blue. What is the
probability that she is wearing matching socks today?
8. Answer:
7
38
8. Solution:
There are
2  2   2  4   2  3  2  9   2 1  1  39 socks
red
blue
gray white orange purple
The probability of drawing a blue is
39-1 =38 socks left in drawer after she takes a blue
8-1=7 blue socks left in drawer
7
38
T1.
Define na ! for n and a positive to be
na !  n  n  a  n  2a  n  3a  ...  n  ka  , where k is the greatest integer for which n  ka.
Then the quotient
728 !
 a b . Find a  b.
182 !
T1. Answer: 13
T1. Solution:
9
728 ! 72  64  56  48 ... 16 8 8  9  8  7  ...  2 1

 9
 49 ,
182 ! 18 16 14 12  ...  4  2 
2  9  8  7  ...  2 1
a  b  13
OHMIO Ciphering 2007 Questions and Solutions
T2.
. In the following equation, each of the letters represents uniquely a different digit in base ten:
YE    ME   TTT
The sum E  T  Y  ?
Answer: 21
Solution:
YE    ME   T 111  T  3 37  ,
therefore, either YE or ME contains a factor of 37. It must be 37
and not 74, because even 74  14 is larger than 999. If E  7, then T  9.
Y 3
M 2
E7
and T  9,
E  M  T  Y  21
999
 27 and
37