# PascalCayleyFermat Electronic Workshop 2001-2 Problem Set 1

Document Sample

```					Pascal/Cayley/Fermat Electronic Workshop 2001-2: Problem Set 1
Posting Dates:     Problems      November 01, 2001
Solutions     December 01, 2001

Part A
( 3
)
1. 4 + 1 × 5 is equal to
5   4
(A)     1
5
(B)         9
1
(C)         2
19
(D)         16
31
(E)         36
Solution:
Evaluating the expression,
(4 + 5 × 4 = 20 + 20 × 4
3 1
) 5   15
( 4  5
)
= 19 × 5
20   4
= 19
16

2. The ratio 1 : 8 is equivalent to
2
(A)     1:4
(B)     1:2
(C)     16 : 1
(D)     4:1
(E)     1 : 16
Solution
We can multiply both sides of a ratio by the same number and preserve the ratio.
Therefore, 1 : 8 is equivalent to 2 × 1 : 2 × 8 or 1 : 16.
2                          2

3. The average of five numbers is 94. If the smallest of these numbers is 86, what is the
average of the remaining four numbers?
(A)     177
(B)     96
(C)     384
(D)     92
(E)     98.5
Solution
Since the average of the five numbers is 94, then the sum of the five numbers is
5 × 94 = 470 . When we remove the number 86, the sum of the remaining numbers is
470 − 86 = 384 , and so the average of the four remaining numbers is 384 = 96.
4
4. The numbers in the sequence 4, 10, 16, 22, 28, … increase by sixes. The numbers in
the sequence 7, 18, 29, 40, 51, … increase by elevens. The number 40 occurs in both
sequences. What is the sum of the next two numbers that occur in both sequences?
(A)    258
(B)    146
(C)    268
(D)    172
(E)    278
Solution
Since 6 and 11 (the differences in the two sequences) have no common factors, the next
common term in the two sequences will be 40 + (6 × 11) = 106 , since the difference
between common terms will be 66. Therefore, the next common term after this is 172,
and the sum of these two common terms is 278.
Part B
5. The smallest positive integer N so that N + 4 is divisible by 5, N + 2 is divisible by
4, and N + 1 is divisible by 3 is
(A)    6
(B)    21
(C)    26
(D)    7
(E)    60
Solution
We use the “check the answers” method.
(A)     6 + 1 is not divisible by 3.
(B)     21 + 2 is not divisible by 4.
(C)    26 satisfies the three requirements.
(D)     7 + 1 is not divisible by 3.
(E)     60 + 1 is not divisible by 3.
Therefore, N = 26 .

6. The smallest positive real number that when added to four times its reciprocal gives
25
6 is
1
(A)     2
6
(B)     25
8
(C)     3
24
(D)     25
3
(E)    2
Solution
Let the number be x. Then we want to solve the equation
4 25
x+      =
x 6
2
x + 4 25
=
x    6
2
6x + 24 = 25x
2
6x − 25x + 24 = 0
(3x − 8 )(2x − 3) = 0
8
Therefore, x =    3   or x = 3 . Since we want the smallest positive value for x, we want
2
x= 3.
2

7. If b = 2 a , c = a + b , and a + 2b + 3c = 70 , then a is equal to
(A)       0
(B)       5
(C)       14
35
(D)        3
(F)       7
Solution
Since b = 2 a , then c = a + b = a + 2 a = 3a . Substituting into the third equation,
a + 2b + 3c = 70
a + 4a + 9a = 70
14a = 70
a=5
8. How many 3-digit positive integers contain the digit 3?
(A)     252
(B)     171
(C)     100
(D)     244
(E)     180
Solution
There are 100 positive integers of the form 3xy.
Next, how many positive integers have 3 as the middle digit, but do not start with a 3 (we
don’t want to double count)? These are numbers of the form x3y, where x is not 3. So
there are 8 possibilities for x (1 through 9, excluding 3) and 10 possibilities for y, so there
are 80 such numbers.
Lastly, how many positive integers end in a 3 and have no other 3’s? These are numbers
of the form xy3, where neither x nor y equals 3. So there are 8 possibilities for x, 9
possibilities for y, and hence 72 possibilities overall.
Combining the three sets, we have 100 + 80 + 72 = 252 possbilities.

[Alternatively, we could note that there are 900 three-digit positive integers, and that to
get the number of these integers containing a 3, we subtract the number of these integers
that do not contain a 3, ie. those numbers of the form xyz, where none of x, y or z is 3.
Thus there are 8 possibilities for x, and possibilities for each of y and z, for a total of
8 × 9 × 9 = 648 of these integers. Therefore, there are 900 − 648 = 252 three-digit
integers that do not contain the integer 3.]

Part C
9. In rectangle ACDE, AC = 38 and AE = 22 . If B and F are the midpoints of AC and
AE respectively, then the area of quadrilateral ABDF is                     B
(A)     209                                             A                            C
(B)     627
(C)     836                                             F
(D)     418
(E)     408
Solution                                                       E
D
Since B is the midpoint of AC, then BC = 19 . Since F is the midpoint of AE, then
FE = 11. Therefore, the area of triangle FED is 1 (38 )(11) = 209 and the area of triangle
2
BCD is 1 (22 )(19) = 209. So the area of quadrilateral ABDF is equal to the area of
2
rectangle ACDE minus the areas of the two triangles, or
38(22) − 2(209 ) = 836 − 418 = 418.

10. The number of positive even factors of 8712 is
(A)   27
(B)   36
(C)   23
(D)     9
(E)     28
Solution
We find the prime factorization of 8712:
8712 = 2 × 4356
= 2 × 2 × 2178
= 2 × 2 × 2 × 1089
= 2 × 2 × 2 × 3× 363
= 2 × 2 × 2 × 3× 3 × 121
= 2 × 2 × 2 × 3× 3 × 11× 11
= 23 32112
Alternatively, we would be more clever and notice
8712 = 8800 − 88
= 88(100 − 1)
= 2311(99 )
= 23 32112
Now an even factor of 8712 must have at least 1 factor of 2 in it, so it could have 1, 2 or 3
factors of 2, as well as 0, 1 or 2 factors of each of 3 and 11.
So we have 3 choices for the number of factors of 2, and 3 choices for the number of
factors of 3, and 3 choices for the number of factors of 11. So overall, there are
3× 3 × 3 = 27 possibilities for an even factor of 8712.