(
a
+
b
)
+
(
c
+
d
)
=
(
a
+
d
)
+
(
b
+
c
)
{\displaystyle (a+b)+(c+d)=(a+d)+(b+c)\,}
a
+
b
+
c
+
d
=
(
a
+
d
)
+
(
b
+
c
)
{\displaystyle a+b+c+d=(a+d)+(b+c)\,}
by associativity
a
+
b
+
d
+
c
=
(
a
+
d
)
+
(
b
+
c
)
{\displaystyle a+b+d+c=(a+d)+(b+c)\,}
by commutativity
a
+
d
+
b
+
c
=
(
a
+
d
)
+
(
b
+
c
)
{\displaystyle a+d+b+c=(a+d)+(b+c)\,}
by commutativity
(
a
+
d
)
+
(
b
+
c
)
=
(
a
+
d
)
+
(
b
+
c
)
{\displaystyle (a+d)+(b+c)=(a+d)+(b+c)\,}
by associativity
(
a
+
b
)
+
(
c
+
d
)
=
(
a
+
d
)
+
(
b
+
c
)
{\displaystyle (a+b)+(c+d)=(a+d)+(b+c)\,}
=
a
+
(
b
+
(
c
+
d
)
)
{\displaystyle =a+(b+(c+d))\,}
by associativity
=
a
+
(
b
+
(
d
+
c
)
)
{\displaystyle =a+(b+(d+c))\,}
by commutativity
=
a
+
(
d
+
(
b
+
c
)
)
{\displaystyle =a+(d+(b+c))\,}
by associativity
=
a
+
(
d
+
(
b
+
c
)
)
{\displaystyle =a+(d+(b+c))\,}
by associativity
(
a
+
b
)
+
(
c
+
d
)
=
(
a
+
c
)
+
(
b
+
d
)
{\displaystyle (a+b)+(c+d)=(a+c)+(b+d)\,}
a
+
b
+
c
+
d
=
(
a
+
c
)
+
(
b
+
d
)
{\displaystyle a+b+c+d=(a+c)+(b+d)\,}
by associativity
a
+
c
+
b
+
d
=
(
a
+
c
)
+
(
b
+
d
)
{\displaystyle a+c+b+d=(a+c)+(b+d)\,}
by commutativity
(
a
+
c
)
+
(
b
+
d
)
=
(
a
+
c
)
+
(
b
+
d
)
{\displaystyle (a+c)+(b+d)=(a+c)+(b+d)\,}
by associativity
(
a
+
b
)
+
(
c
+
d
)
=
(
a
+
c
)
+
(
b
+
d
)
{\displaystyle (a+b)+(c+d)=(a+c)+(b+d)\,}
a
+
(
b
+
(
c
+
d
)
)
=
(
a
+
c
)
+
(
b
+
d
)
{\displaystyle a+(b+(c+d))=(a+c)+(b+d)\,}
by associativity
a
+
(
(
b
+
c
)
+
d
)
=
(
a
+
c
)
+
(
b
+
d
)
{\displaystyle a+((b+c)+d)=(a+c)+(b+d)\,}
by associativity
a
+
(
(
c
+
b
)
+
d
)
=
(
a
+
c
)
+
(
b
+
d
)
{\displaystyle a+((c+b)+d)=(a+c)+(b+d)\,}
by commutativity
a
+
(
c
+
(
b
+
d
)
)
=
(
a
+
c
)
+
(
b
+
d
)
{\displaystyle a+(c+(b+d))=(a+c)+(b+d)\,}
by associativity
(
a
+
c
)
+
(
b
+
d
)
=
(
a
+
c
)
+
(
b
+
d
)
{\displaystyle (a+c)+(b+d)=(a+c)+(b+d)\,}
by associativity
(
a
−
b
)
+
(
c
−
d
)
=
(
a
+
c
)
+
(
−
b
−
d
)
{\displaystyle (a-b)+(c-d)=(a+c)+(-b-d)\,}
a
−
b
+
c
−
d
=
(
a
+
c
)
+
(
−
b
−
d
)
{\displaystyle a-b+c-d=(a+c)+(-b-d)\,}
by associativity
a
+
c
−
b
−
d
=
(
a
+
c
)
+
(
−
b
−
d
)
{\displaystyle a+c-b-d=(a+c)+(-b-d)\,}
by commutativity
(
a
+
c
)
+
(
−
b
−
d
)
=
(
a
+
c
)
+
(
−
b
−
d
)
{\displaystyle (a+c)+(-b-d)=(a+c)+(-b-d)\,}
by associativity
−
2
+
x
=
4
{\displaystyle -2+x=4\,}
x
=
4
+
2
{\displaystyle x=4+2\,}
x
=
6
{\displaystyle x=6\,}
2
−
x
=
5
{\displaystyle 2-x=5\,}
−
x
=
3
{\displaystyle -x=3\,}
x
=
−
3
{\displaystyle x=-3\,}
multiply both sides by -1.
Although these can be done by hand the exercises suggest that you derive a general formula for finding the final population figures. The following one will suffice for the stated problems, where
p
{\displaystyle p\,}
x
{\displaystyle x\,}
y
1
{\displaystyle y_{1}\,}
y
2
{\displaystyle y_{2}\,}
n
{\displaystyle n\,}
x
{\displaystyle x\,}
p
⋅
x
(
y
2
−
y
1
n
)
{\displaystyle p\cdot x^{({\frac {y_{2}-y_{1}}{n}})}}
200000
⋅
3
(
2215
−
1915
50
)
=
200000
⋅
3
6
=
145800000
{\displaystyle 200000\cdot 3^{({\frac {2215-1915}{50}})}=200000\cdot 3^{6}=145800000}
Write out the powers of 2 not exceeding the largest number in the problem set. Systematically check to see if each power divides the given number, starting with the largest.
Write out the powers of 3 not exceeding the largest number in the problem set. Systematically check to see if each power divides the given number, starting with the largest.
Examining the definitions first, if
a
≡
b
(
mod
5
)
{\displaystyle a\equiv b{\pmod {5}}}
x
≡
y
(
mod
5
)
{\displaystyle x\equiv y{\pmod {5}}}
a
−
b
=
5
n
{\displaystyle a-b=5n\,}
x
−
y
=
5
m
{\displaystyle x-y=5m\,}
a
+
x
≡
b
+
y
(
mod
5
)
{\displaystyle a+x\equiv b+y{\pmod {5}}}
(
a
+
x
)
−
(
b
+
y
)
=
5
k
{\displaystyle (a+x)-(b+y)=5k\,}
(
a
+
x
)
−
(
b
+
y
)
=
(
a
−
b
)
+
(
x
−
y
)
{\displaystyle (a+x)-(b+y)=(a-b)+(x-y)\,}
(
a
+
x
)
−
(
b
+
y
)
=
5
n
+
5
m
{\displaystyle (a+x)-(b+y)=5n+5m\,}
(
a
+
x
)
−
(
b
+
y
)
=
5
(
n
+
m
)
{\displaystyle (a+x)-(b+y)=5(n+m)\,}
k
=
(
n
+
m
)
{\displaystyle k=(n+m)\,}
(
a
+
x
)
−
(
b
+
y
)
=
5
k
{\displaystyle (a+x)-(b+y)=5k\,}
Examine the definitions in part a) again. We have
a
−
b
=
5
n
{\displaystyle a-b=5n\,}
x
−
y
=
5
m
{\displaystyle x-y=5m\,}
a
{\displaystyle a\,}
x
{\displaystyle x\,}
a
=
5
n
+
b
{\displaystyle a=5n+b\,}
x
=
5
m
+
y
{\displaystyle x=5m+y\,}
a
x
=
(
5
n
+
b
)
(
5
m
+
y
)
{\displaystyle ax=(5n+b)(5m+y)\,}
a
x
=
25
n
m
+
5
n
y
+
5
m
b
+
b
y
{\displaystyle ax=25nm+5ny+5mb+by\,}
a
x
=
5
(
5
n
m
+
n
y
+
m
b
)
+
b
y
{\displaystyle ax=5(5nm+ny+mb)+by\,}
t
=
5
n
m
+
n
y
+
m
b
{\displaystyle t=5nm+ny+mb\,}
a
x
=
5
t
+
b
y
{\displaystyle ax=5t+by\,}
a
x
−
b
y
=
5
t
+
b
y
−
b
y
{\displaystyle ax-by=5t+by-by\,}
a
x
−
b
y
=
5
t
{\displaystyle ax-by=5t\,}
120, 720, 5040 and 40320.
(
m
n
)
=
(
m
m
−
n
)
{\displaystyle {\binom {m}{n}}={\binom {m}{m-n}}}
m
!
n
!
(
m
−
n
)
!
=
m
!
(
m
−
n
)
!
(
m
−
(
m
−
n
)
)
!
{\displaystyle {\frac {m!}{n!(m-n)!}}={\frac {m!}{(m-n)!(m-(m-n))!}}}
m
!
n
!
(
m
−
n
)
!
=
m
!
(
m
−
n
)
!
(
m
−
m
+
n
)
!
{\displaystyle {\frac {m!}{n!(m-n)!}}={\frac {m!}{(m-n)!(m-m+n)!}}}
m
!
n
!
(
m
−
n
)
!
=
m
!
(
m
−
n
)
!
n
!
{\displaystyle {\frac {m!}{n!(m-n)!}}={\frac {m!}{(m-n)!n!}}}
(
m
n
)
+
(
m
n
−
1
)
=
(
m
+
1
n
)
{\displaystyle {\binom {m}{n}}+{\binom {m}{n-1}}={\binom {m+1}{n}}}
m
!
n
!
(
m
−
n
)
!
+
m
!
(
n
−
1
)
!
(
m
−
n
+
1
)
!
=
(
m
+
1
)
!
n
!
(
m
−
n
+
1
)
!
{\displaystyle {\frac {m!}{n!(m-n)!}}+{\frac {m!}{(n-1)!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
n
n
{\displaystyle {\frac {n}{n}}}
(
m
−
n
+
1
)
(
m
−
n
+
1
)
{\displaystyle {\frac {(m-n+1)}{(m-n+1)}}}
n
!
(
m
−
n
+
1
)
!
{\displaystyle n!(m-n+1)!}
1
⋅
(
m
−
n
+
1
)
(
m
−
n
+
1
)
⋅
m
!
n
!
(
m
−
n
)
!
+
n
n
⋅
1
⋅
m
!
(
n
−
1
)
!
(
m
−
n
+
1
)
!
=
1
⋅
1
⋅
(
m
+
1
)
!
n
!
(
m
−
n
+
1
)
!
{\displaystyle 1\cdot {\frac {(m-n+1)}{(m-n+1)}}\cdot {\frac {m!}{n!(m-n)!}}+{\frac {n}{n}}\cdot 1\cdot {\frac {m!}{(n-1)!(m-n+1)!}}=1\cdot 1\cdot {\frac {(m+1)!}{n!(m-n+1)!}}}
a
!
(
a
+
1
)
=
(
a
+
1
)
!
{\displaystyle a!(a+1)=(a+1)!}
a
(
a
−
1
)
!
=
a
!
{\displaystyle a(a-1)!=a!}
m
!
(
m
−
n
+
1
)
n
!
(
m
−
n
+
1
)
!
+
m
!
n
n
!
(
m
−
n
+
1
)
!
=
(
m
+
1
)
!
n
!
(
m
−
n
+
1
)
!
{\displaystyle {\frac {m!(m-n+1)}{n!(m-n+1)!}}+{\frac {m!n}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
m
!
(
m
−
n
+
1
)
+
m
!
n
n
!
(
m
−
n
+
1
)
!
=
(
m
+
1
)
!
n
!
(
m
−
n
+
1
)
!
{\displaystyle {\frac {m!(m-n+1)+m!n}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
m
!
(
(
m
−
n
+
1
)
+
n
)
n
!
(
m
−
n
+
1
)
!
=
(
m
+
1
)
!
n
!
(
m
−
n
+
1
)
!
{\displaystyle {\frac {m!((m-n+1)+n)}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
m
!
(
m
+
1
)
n
!
(
m
−
n
+
1
)
!
=
(
m
+
1
)
!
n
!
(
m
−
n
+
1
)
!
{\displaystyle {\frac {m!(m+1)}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
(
m
+
1
)
!
n
!
(
m
−
n
+
1
)
!
=
(
m
+
1
)
!
n
!
(
m
−
n
+
1
)
!
{\displaystyle {\frac {(m+1)!}{n!(m-n+1)!}}={\frac {(m+1)!}{n!(m-n+1)!}}}
(
m
n
)
+
(
m
n
−
1
)
=
(
m
+
1
n
)
{\displaystyle {\binom {m}{n}}+{\binom {m}{n-1}}={\binom {m+1}{n}}}
2
x
−
1
3
x
+
2
=
7
{\displaystyle {\frac {2x-1}{3x+2}}=7\,}
2
x
−
1
=
7
(
3
x
+
2
)
{\displaystyle 2x-1=7(3x+2)\,}
2
x
−
1
=
21
x
+
14
{\displaystyle 2x-1=21x+14\,}
−
19
x
=
15
{\displaystyle -19x=15\,}
x
=
−
15
19
{\displaystyle x=-{\frac {15}{19}}\,}
1
x
+
y
−
1
x
−
y
=
−
2
y
x
2
−
y
2
{\displaystyle {\frac {1}{x+y}}-{\frac {1}{x-y}}={\frac {-2y}{x^{2}-y^{2}}}}
(
x
−
y
)
−
(
x
+
y
)
(
x
+
y
)
(
x
−
y
)
=
.
.
.
{\displaystyle {\frac {(x-y)-(x+y)}{(x+y)(x-y)}}=...}
x
+
(
−
x
)
+
(
−
y
)
+
(
−
y
)
x
2
+
y
x
−
y
x
−
y
2
=
.
.
.
{\displaystyle {\frac {x+(-x)+(-y)+(-y)}{x^{2}+yx-yx-y^{2}}}=...}
−
2
y
x
2
−
y
2
=
−
2
y
x
2
−
y
2
{\displaystyle {\frac {-2y}{x^{2}-y^{2}}}={\frac {-2y}{x^{2}-y^{2}}}}
x
3
−
1
x
−
1
=
1
+
x
+
x
2
{\displaystyle {\frac {x^{3}-1}{x-1}}=1+x+x^{2}}
Recall from the chapter that if
a
b
=
c
d
{\displaystyle {\frac {a}{b}}={\frac {c}{d}}}
a
d
=
b
c
{\displaystyle ad=bc}
1
⋅
(
x
3
−
1
)
=
(
x
−
1
)
(
1
+
x
+
x
2
)
{\displaystyle 1\cdot (x^{3}-1)=(x-1)(1+x+x^{2})}
x
3
−
1
=
x
+
x
2
+
x
3
−
1
−
x
−
x
2
{\displaystyle x^{3}-1=x+x^{2}+x^{3}-1-x-x^{2}\,}
x
3
−
1
=
x
3
−
1
{\displaystyle x^{3}-1=x^{3}-1\,}
