DISINI 10-11
UNIVERSITY OF NORTHERN COLORADO
MATHEMATICS CONTEST
First Round
For all Colorado Students Grades 7-12
November 6, 2010
You have 90 minutes – no calculators allowed
•
•
An arithmetic sequence with common difference d and first term a is:
!, ! + !, ! + 2!, ! + 3!, …
The perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, …
1. There are five “lines” of three circles each
in the diagram. Insert the numbers
1, 2, 3, 4, 5, 6, 7, one in each circle, so
that the sum of the three numbers
in each line is the same. Which
number must be in the top most circle?
2. Five of the ten discs are dark and five are light;
and they alternate as shown in the pattern. If you are only allowed to interchange the positions of
the two neighboring discs in a single move, what is the least number of moves required to get all
the dark discs together at the right hand side?
3. ABCD is a square with side length 8.
E and F are midpoints of sides. ED and FC
are radii of quarter circles centered at D and
C respectively. EF is the diameter of the
upper semicircle. What is the area of the
shaded, ice cream cone, area?
4. The four numbers ! − 3, ! − 1, ! + 1, ! + 3 are in arithmetic progression with common difference
! = 2. Show that their product plus 16 is a perfect square. Specifically, find A (in terms of a) so
that ! − 3 ! − 1 ! + 1 ! + 3 + 16 = !! .
5. An isosceles trapezoid has sides
15, 28, 15, and 52 as shown. Find the
length of a diagonal.
6. A sequence of numbers is generated by the following initial condition and recursion:
!! = 3
!!!! =
!!
!!!!
(a) Find !!
(b) Find !!"#
7. Consider the arrangement (to the right) of the
integers greater than 1.
(a) In which column will 100 fall?
(b) In which column will 1000 fall?
(c) In which column will 2011 fall?
A
B
2
9 8
10
17 16
⋮
C
3
7
11
15
⋮
D E
4 5
6
12 13
14
⋮
8. Determine the 2010th term of the following sequence:
1, 2, 3, 6, 7, 14, 15, 30, 31, 62, 63, …
where a term in an even numbered spot is twice the previous term, and a term in an odd numbered
spot is one more than the previous term.
9. Five positive integers a, b, c, d, and e are written on the blackboard. When you add them four at
time the sums are 89, 99, 106, 110 and 112. What is the value of the smallest of the integers a, b, c,
d, e?
10. Let !(!) be the number of ways of selecting three distinct integers from 1,2,3, … , ! so that they
are the lengths of the sides of a triangle. For example, ! 5 = 3; the only possibilities are
2-3-4, 2-4-5, and 3-4-5. Find ! 7 and ! 8 .
Brief Solutions – First Round
November 8, 2010
1.
4; Since the top number is used three times and the other six are used twice, the top number must be 4, the
middle number of the entries 1, 2, 3, 4, 5, 6, 7.
2.
15; Interchanging one at a time starting at the right requires 1 + 2 + 3 + 4 + 5 = 15 swaps.
3.
32; From the picture the shaded area is 2! + 2!
or half the area of the 8x8 square.
4.
! ! − 5; ! − 3 ! − 1 ! + 1 ! + 3 + 16 = ! ! − 9 ! ! − 1 + 16 = ! ! − 10! ! + 25 = ! ! − 5
5.
41; The altitude is 9.
!
The diagonal d is the
hypotenuse of the right
triangle whose legs are 9 and 40.
Then ! = 81 + 1600 = 1681 = 41
6.
(a) !! =
!!
!!!!
=
!!
!
.
(b) To find !!"# , start generating the sequence. !! =
!!
!!!!
!
!!
!
!!!!
= − , !! =
= 3, !! =
!!
!!!!
!
= − , the
!
!
same as !! . This sequence repeats in cycles of length three. Since 501 is a multiple of 3, !!"# = − .
!
7.
(a) 100 is in column D, as part of the arithmetic subsequence 4, 12, 20, …
You can solve 4 + 8! = 100 for ! = 12.
(b) 1000 is in column B as part of the subsequence 8, 16, 24, 32, …
You can solve 8 + 8! = 1000 for ! = 124.
(c) 2011 is in column C; 3 + 4! = 2011 gives ! = 502.
8.
2!""# − 2; Make a table
Term #
1
3
5
7
9
11
Term
1
3
7
15
31
63 = 2! − 1
Since 2! − 1 is the 11th term, the 2011th term would be 2!""# − 1. Then 2!""# − 2 would be the 2010th
term.
9.
17; ! + ! + ! + ! = 89
! + ! + ! + ! = 99
! + ! + ! + ! = 106
! + ! + ! + ! = 110
! + ! + ! + ! = 112
Adding both sides, 4! + 4! + 4! + 4! + 4! = 516 and ! + ! + ! + ! + ! = 129. Since 112 is the
largest, ! = 129 − 112 = 17 is the smallest.
10. ! 7 = 13 and ! 8 = 22
University of Northern Colorado
MATHEMATICS CONTEST
FINAL ROUND January 29, 2011
For Colorado Students Grades 7-12
1.
•
!!, read as n factorial, is computed as !! = 1 ∙ 2 ∙ 3 ∙ 4 ∙∙∙ !
•
The factorials are 1, 2, 6, 24, 120, 720, …
•
The square integers are 1, 4, 9, 16, 25, 36, 49, 64, 81, …
The largest integer n so that 3! evenly divides 9! = 1 ∙ 2 ∙ 3 ∙ 4 ∙ 5 ∙ 6 ∙ 7 ∙ 8 ∙ 9 is ! = 4. Determine
the largest integer n so that 3! evenly divides 85! = 1 ∙ 2 ∙ 3 ∙ 4 ⋯ 84 ∙ 85.
2.
Let m and n be positive integers. List all the integers in the set
20 ,21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31 that cannot be written in the form ! + ! + !".
As an example, 20 can be so expressed since 20 = 2 + 6 + 2 ∙ 6.
3.
The two congruent rectangles shown have
dimensions 5 in. by 25 in. What is the area
of the shaded overlap region?
4.
Let ! = 2, 5, 10, 17, ⋯ , !! + 1, ⋯
be the set of all positive squares plus 1 and
! = 101, 104, 109, 116, ⋯ , !! + 100, ⋯ be the set of all positive squares plus 100.
(a) What is the smallest number in both A and B?
(b) Find all numbers that are in both A and B.
5.
Determine the area of the square ABCD,
with the given lengths along a zigzag
line connecting B and D.
OVER
6. What is the remainder when 1! + 2! + 3! + ⋯ + 2011! is divided by 18?
7. What is the sum of the first 999 terms of the sequence 1, 2, 3, 6, 7, 14, 15, 30, 31, 62, 63, ⋯ that appeared
on the First Round? Recall that a term in an even numbered position is twice the previous term, while a
term in an odd numbered position is one more that the previous term.
8. The integer 45 can be expressed as a sum of two squares as 45 = 3! + 6! .
(a) Express 74 as the sum of two squares.
(b) Express the product 45 ∙ 74 as the sum of two squares.
(c) Prove that the product of two sums of two squares, !! + ! ! ! ! + ! ! , can be represented
as the sum of two squares.
9. Let !(!) be the number of ways of selecting three distinct numbers from 1, 2, 3, ⋯ , ! so that they are
the lengths of the sides of a triangle. As an example, ! 5 = 3; the only possibilities are 2-3-4, 2-4-5,
and 3-4-5.
(a) Determine a recursion for !(!).
(b) Determine a closed formula for !(!).
10. The integers 1, 2, 3, ⋯ , 50 are written on the blackboard. Select any two, call them m and n and replace
these two with the one number ! + ! + !". Continue doing this until only one number remains and
explain, with proof, what happens. Also explain with proof what happens in general as you replace 50
with n. As an example, if you select 3 and 17 you replace them with 3 + 17 + 51 = 71. If you select 5
and 7, replace them with 47. You now have two 47’s in this case but that’s OK.
11. Tie breaker – Generalize problem #2, and prove your statement.
Brief Solutions Final Round
January 29, 2011
1. n=41; Look at 1 ∙ 2 ∙ 3 ∙ 4 ∙ 5 ∙ 6 ⋯ 81 ∙ 82 ∙ 83 ∙ 84 ∙ 85 closely. Every 3rd one is divisible by 3 (there are 28 of
these); every 9th one has an extra 3 (there are 9 of these) and every 27th one has another extra 3 (there are 3 of
these, namely 27, 54, and 81); and then add 1 for 81.
2. 22, 28 and 30; One way is to test them starting with 2 + 6 + 12 = 20. Alternatively, notice that
! + ! + !" = ! + 1 ! + 1 − 1. Call this number Q. Then ! + 1 is not a prime since each of ! + 1, ! + 1
are greater than 2. So those that do not work are one less than a prime. The only primes in the given set are 23, 29
and 31.
3. 65; ! ! = 25 + 25 − !
!
= 650 − 50! + ! ! ;
650 = 50!; ! = 13. Area=5x13=65
4. (a) 101
(b) 101, 325, 2501; !! + 1 = ! ! + 100 implies ! − ! ! + ! = 3 ∙ 3 ∙ 11. Match up factors:
! − ! = 1 !"# ! + ! = 99 gives ! = 50, ! = 49 and !! + 1 = 2501; ! − ! = 3, ! + ! = 33
gives ! = 18, ! = 15 and !! + 1 = 325; ! − ! = 9, ! + ! = 11 gives ! = 10, ! = 1 and !! + 1 = 101.
5. 160;
!
!
=
!"
!!!
gives ! = 3. Then !!! = 45, !!! = 125,
!
d1+d2=8 5. If y is the side length, ! ! + ! ! = 8 5
and ! ! = 160, the area.
Alternate solution to #5. Extend drawing as shown.
Then compute the diagonal of the square to be 8 5 .
6. 9; 6! + 7! + ⋯ + 2011! is divisible by 18. The remainder when 1! + 2! + 3! + 4! + 5! = 153 is divided
by 18 is 9.
7. 2!"# − 1504; Split the sequence into two sequences: 1 + 3 + 7 + 15 + ⋯ + 2!"" − 1 and
2 + 6 + 14 + 30 + ⋯ + 2!"" − 2 . Then 2! − 1 + 2! − 1 + ⋯ + 2!"" − 1 = 2!"# − 502 and
2! − 2 + 2! − 2 + ⋯ + 2!"" − 2 = 2!"# − 1002. Now add to get
2!"# − 502 + 2!"# − 1002 = 2 ∙ 2!"# − 1504 = 2!"# − 1504
8. (a) 74 = 5! + 7!
(b) 45 ∙ 74 = 3! + 6! 5! + 7! = 3 ∙ 5 + 6 ∙ 7
!
+ 6∙5−3∙7
!
= 57! + 9! . Another correct answer;
51! + 27! .
(c) ! ! + ! ! ! ! + ! ! = !" + !"
!
+ !" − !"
!
9. (a) First collect data, carefully, and make a table:
n
3
4
5
6
7
8
9
10
11
T(n)
0
1
3
7
13
22
34
50
70
If you add two consecutive terms you get 1, 4, 10, 20, 35, 56, 84, 120, the entries on the 4th diagonal in
!
Pascal’s Triangle. Hence ! ! + 1 + ! ! =
is a recursion for ! ! .
3
(b) Iteration of the recursion gives ! ! =
!−1
!−2
!−3
!−4
−
+
−
+⋯
3
3
3
3
The use of generating functions will also show this.
Alternate recursions and formulas include:
! ! =! !−1 +
!!! !
!"# ! !"!#
!
!!! !!!
!"# ! !""
!
!!! !!! !!!!
! ! =
!"
!!! !!! !!!!
!"
! ! =
=
!
!
!
!
!"# ! !""
+
!
!
−1
!
=
! !!! !!!!
!"
!"# ! !"!#
! − 2 ! − 3 + ! − 4 ! − 5 + ⋯ + 3 ∙ 2 !ℎ!" ! !" !""
! − 2 ! − 3 + ! − 4 ! − 5 ⋯ + 2 ∙ 1 !ℎ!" ! !" !"!#.
10. First try 1, 2, 3, … , ! for ! = 2, 3, 4, 5. The crossing off process yields 5, 23, 119, 719 each one being one less
than a factorial. So for general n you should end up with ! + 1 ! − 1. Now look at n=3 again and replace 1, 2, 3
with a, b, c (order does not matter). Crossing off gives you
! + ! + !" + ! + ! + ! + !" ! = ! + ! + ! + !" + !" + !" + !"#, reminding one of the coefficients in
! − ! ! − ! ! − ! = ! ! − ! + ! + ! ! ! + !" + !" + !" ! − !"#. Now let ! = −1, and watch what
happens remember that !, !, ! = 1,2,3 .
There are other approaches.
11. See solution to #2. Integers that are one less than a prime cannot be written in the form ! + ! + !.
MATHEMATICS CONTEST
First Round
For all Colorado Students Grades 7-12
November 6, 2010
You have 90 minutes – no calculators allowed
•
•
An arithmetic sequence with common difference d and first term a is:
!, ! + !, ! + 2!, ! + 3!, …
The perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, …
1. There are five “lines” of three circles each
in the diagram. Insert the numbers
1, 2, 3, 4, 5, 6, 7, one in each circle, so
that the sum of the three numbers
in each line is the same. Which
number must be in the top most circle?
2. Five of the ten discs are dark and five are light;
and they alternate as shown in the pattern. If you are only allowed to interchange the positions of
the two neighboring discs in a single move, what is the least number of moves required to get all
the dark discs together at the right hand side?
3. ABCD is a square with side length 8.
E and F are midpoints of sides. ED and FC
are radii of quarter circles centered at D and
C respectively. EF is the diameter of the
upper semicircle. What is the area of the
shaded, ice cream cone, area?
4. The four numbers ! − 3, ! − 1, ! + 1, ! + 3 are in arithmetic progression with common difference
! = 2. Show that their product plus 16 is a perfect square. Specifically, find A (in terms of a) so
that ! − 3 ! − 1 ! + 1 ! + 3 + 16 = !! .
5. An isosceles trapezoid has sides
15, 28, 15, and 52 as shown. Find the
length of a diagonal.
6. A sequence of numbers is generated by the following initial condition and recursion:
!! = 3
!!!! =
!!
!!!!
(a) Find !!
(b) Find !!"#
7. Consider the arrangement (to the right) of the
integers greater than 1.
(a) In which column will 100 fall?
(b) In which column will 1000 fall?
(c) In which column will 2011 fall?
A
B
2
9 8
10
17 16
⋮
C
3
7
11
15
⋮
D E
4 5
6
12 13
14
⋮
8. Determine the 2010th term of the following sequence:
1, 2, 3, 6, 7, 14, 15, 30, 31, 62, 63, …
where a term in an even numbered spot is twice the previous term, and a term in an odd numbered
spot is one more than the previous term.
9. Five positive integers a, b, c, d, and e are written on the blackboard. When you add them four at
time the sums are 89, 99, 106, 110 and 112. What is the value of the smallest of the integers a, b, c,
d, e?
10. Let !(!) be the number of ways of selecting three distinct integers from 1,2,3, … , ! so that they
are the lengths of the sides of a triangle. For example, ! 5 = 3; the only possibilities are
2-3-4, 2-4-5, and 3-4-5. Find ! 7 and ! 8 .
Brief Solutions – First Round
November 8, 2010
1.
4; Since the top number is used three times and the other six are used twice, the top number must be 4, the
middle number of the entries 1, 2, 3, 4, 5, 6, 7.
2.
15; Interchanging one at a time starting at the right requires 1 + 2 + 3 + 4 + 5 = 15 swaps.
3.
32; From the picture the shaded area is 2! + 2!
or half the area of the 8x8 square.
4.
! ! − 5; ! − 3 ! − 1 ! + 1 ! + 3 + 16 = ! ! − 9 ! ! − 1 + 16 = ! ! − 10! ! + 25 = ! ! − 5
5.
41; The altitude is 9.
!
The diagonal d is the
hypotenuse of the right
triangle whose legs are 9 and 40.
Then ! = 81 + 1600 = 1681 = 41
6.
(a) !! =
!!
!!!!
=
!!
!
.
(b) To find !!"# , start generating the sequence. !! =
!!
!!!!
!
!!
!
!!!!
= − , !! =
= 3, !! =
!!
!!!!
!
= − , the
!
!
same as !! . This sequence repeats in cycles of length three. Since 501 is a multiple of 3, !!"# = − .
!
7.
(a) 100 is in column D, as part of the arithmetic subsequence 4, 12, 20, …
You can solve 4 + 8! = 100 for ! = 12.
(b) 1000 is in column B as part of the subsequence 8, 16, 24, 32, …
You can solve 8 + 8! = 1000 for ! = 124.
(c) 2011 is in column C; 3 + 4! = 2011 gives ! = 502.
8.
2!""# − 2; Make a table
Term #
1
3
5
7
9
11
Term
1
3
7
15
31
63 = 2! − 1
Since 2! − 1 is the 11th term, the 2011th term would be 2!""# − 1. Then 2!""# − 2 would be the 2010th
term.
9.
17; ! + ! + ! + ! = 89
! + ! + ! + ! = 99
! + ! + ! + ! = 106
! + ! + ! + ! = 110
! + ! + ! + ! = 112
Adding both sides, 4! + 4! + 4! + 4! + 4! = 516 and ! + ! + ! + ! + ! = 129. Since 112 is the
largest, ! = 129 − 112 = 17 is the smallest.
10. ! 7 = 13 and ! 8 = 22
University of Northern Colorado
MATHEMATICS CONTEST
FINAL ROUND January 29, 2011
For Colorado Students Grades 7-12
1.
•
!!, read as n factorial, is computed as !! = 1 ∙ 2 ∙ 3 ∙ 4 ∙∙∙ !
•
The factorials are 1, 2, 6, 24, 120, 720, …
•
The square integers are 1, 4, 9, 16, 25, 36, 49, 64, 81, …
The largest integer n so that 3! evenly divides 9! = 1 ∙ 2 ∙ 3 ∙ 4 ∙ 5 ∙ 6 ∙ 7 ∙ 8 ∙ 9 is ! = 4. Determine
the largest integer n so that 3! evenly divides 85! = 1 ∙ 2 ∙ 3 ∙ 4 ⋯ 84 ∙ 85.
2.
Let m and n be positive integers. List all the integers in the set
20 ,21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31 that cannot be written in the form ! + ! + !".
As an example, 20 can be so expressed since 20 = 2 + 6 + 2 ∙ 6.
3.
The two congruent rectangles shown have
dimensions 5 in. by 25 in. What is the area
of the shaded overlap region?
4.
Let ! = 2, 5, 10, 17, ⋯ , !! + 1, ⋯
be the set of all positive squares plus 1 and
! = 101, 104, 109, 116, ⋯ , !! + 100, ⋯ be the set of all positive squares plus 100.
(a) What is the smallest number in both A and B?
(b) Find all numbers that are in both A and B.
5.
Determine the area of the square ABCD,
with the given lengths along a zigzag
line connecting B and D.
OVER
6. What is the remainder when 1! + 2! + 3! + ⋯ + 2011! is divided by 18?
7. What is the sum of the first 999 terms of the sequence 1, 2, 3, 6, 7, 14, 15, 30, 31, 62, 63, ⋯ that appeared
on the First Round? Recall that a term in an even numbered position is twice the previous term, while a
term in an odd numbered position is one more that the previous term.
8. The integer 45 can be expressed as a sum of two squares as 45 = 3! + 6! .
(a) Express 74 as the sum of two squares.
(b) Express the product 45 ∙ 74 as the sum of two squares.
(c) Prove that the product of two sums of two squares, !! + ! ! ! ! + ! ! , can be represented
as the sum of two squares.
9. Let !(!) be the number of ways of selecting three distinct numbers from 1, 2, 3, ⋯ , ! so that they are
the lengths of the sides of a triangle. As an example, ! 5 = 3; the only possibilities are 2-3-4, 2-4-5,
and 3-4-5.
(a) Determine a recursion for !(!).
(b) Determine a closed formula for !(!).
10. The integers 1, 2, 3, ⋯ , 50 are written on the blackboard. Select any two, call them m and n and replace
these two with the one number ! + ! + !". Continue doing this until only one number remains and
explain, with proof, what happens. Also explain with proof what happens in general as you replace 50
with n. As an example, if you select 3 and 17 you replace them with 3 + 17 + 51 = 71. If you select 5
and 7, replace them with 47. You now have two 47’s in this case but that’s OK.
11. Tie breaker – Generalize problem #2, and prove your statement.
Brief Solutions Final Round
January 29, 2011
1. n=41; Look at 1 ∙ 2 ∙ 3 ∙ 4 ∙ 5 ∙ 6 ⋯ 81 ∙ 82 ∙ 83 ∙ 84 ∙ 85 closely. Every 3rd one is divisible by 3 (there are 28 of
these); every 9th one has an extra 3 (there are 9 of these) and every 27th one has another extra 3 (there are 3 of
these, namely 27, 54, and 81); and then add 1 for 81.
2. 22, 28 and 30; One way is to test them starting with 2 + 6 + 12 = 20. Alternatively, notice that
! + ! + !" = ! + 1 ! + 1 − 1. Call this number Q. Then ! + 1 is not a prime since each of ! + 1, ! + 1
are greater than 2. So those that do not work are one less than a prime. The only primes in the given set are 23, 29
and 31.
3. 65; ! ! = 25 + 25 − !
!
= 650 − 50! + ! ! ;
650 = 50!; ! = 13. Area=5x13=65
4. (a) 101
(b) 101, 325, 2501; !! + 1 = ! ! + 100 implies ! − ! ! + ! = 3 ∙ 3 ∙ 11. Match up factors:
! − ! = 1 !"# ! + ! = 99 gives ! = 50, ! = 49 and !! + 1 = 2501; ! − ! = 3, ! + ! = 33
gives ! = 18, ! = 15 and !! + 1 = 325; ! − ! = 9, ! + ! = 11 gives ! = 10, ! = 1 and !! + 1 = 101.
5. 160;
!
!
=
!"
!!!
gives ! = 3. Then !!! = 45, !!! = 125,
!
d1+d2=8 5. If y is the side length, ! ! + ! ! = 8 5
and ! ! = 160, the area.
Alternate solution to #5. Extend drawing as shown.
Then compute the diagonal of the square to be 8 5 .
6. 9; 6! + 7! + ⋯ + 2011! is divisible by 18. The remainder when 1! + 2! + 3! + 4! + 5! = 153 is divided
by 18 is 9.
7. 2!"# − 1504; Split the sequence into two sequences: 1 + 3 + 7 + 15 + ⋯ + 2!"" − 1 and
2 + 6 + 14 + 30 + ⋯ + 2!"" − 2 . Then 2! − 1 + 2! − 1 + ⋯ + 2!"" − 1 = 2!"# − 502 and
2! − 2 + 2! − 2 + ⋯ + 2!"" − 2 = 2!"# − 1002. Now add to get
2!"# − 502 + 2!"# − 1002 = 2 ∙ 2!"# − 1504 = 2!"# − 1504
8. (a) 74 = 5! + 7!
(b) 45 ∙ 74 = 3! + 6! 5! + 7! = 3 ∙ 5 + 6 ∙ 7
!
+ 6∙5−3∙7
!
= 57! + 9! . Another correct answer;
51! + 27! .
(c) ! ! + ! ! ! ! + ! ! = !" + !"
!
+ !" − !"
!
9. (a) First collect data, carefully, and make a table:
n
3
4
5
6
7
8
9
10
11
T(n)
0
1
3
7
13
22
34
50
70
If you add two consecutive terms you get 1, 4, 10, 20, 35, 56, 84, 120, the entries on the 4th diagonal in
!
Pascal’s Triangle. Hence ! ! + 1 + ! ! =
is a recursion for ! ! .
3
(b) Iteration of the recursion gives ! ! =
!−1
!−2
!−3
!−4
−
+
−
+⋯
3
3
3
3
The use of generating functions will also show this.
Alternate recursions and formulas include:
! ! =! !−1 +
!!! !
!"# ! !"!#
!
!!! !!!
!"# ! !""
!
!!! !!! !!!!
! ! =
!"
!!! !!! !!!!
!"
! ! =
=
!
!
!
!
!"# ! !""
+
!
!
−1
!
=
! !!! !!!!
!"
!"# ! !"!#
! − 2 ! − 3 + ! − 4 ! − 5 + ⋯ + 3 ∙ 2 !ℎ!" ! !" !""
! − 2 ! − 3 + ! − 4 ! − 5 ⋯ + 2 ∙ 1 !ℎ!" ! !" !"!#.
10. First try 1, 2, 3, … , ! for ! = 2, 3, 4, 5. The crossing off process yields 5, 23, 119, 719 each one being one less
than a factorial. So for general n you should end up with ! + 1 ! − 1. Now look at n=3 again and replace 1, 2, 3
with a, b, c (order does not matter). Crossing off gives you
! + ! + !" + ! + ! + ! + !" ! = ! + ! + ! + !" + !" + !" + !"#, reminding one of the coefficients in
! − ! ! − ! ! − ! = ! ! − ! + ! + ! ! ! + !" + !" + !" ! − !"#. Now let ! = −1, and watch what
happens remember that !, !, ! = 1,2,3 .
There are other approaches.
11. See solution to #2. Integers that are one less than a prime cannot be written in the form ! + ! + !.