A positive integer is represented, as 21022200 in a base 'n' system, and as 10112121 in a base n+1 system. What is the 'n', that is larger than 2, in the decimal system?
(Note: for example, the decimal number 37 means 3*10+7. 102 in the base 3 system is equivalent to 1*32+0*3+2.)Inspecting a 7th-degree polynomial of 'n' by taking its derivatives
From the given conditions,
2n7+n6+2n4+2n3+2n2
=(n+1)7+(n+1)5+(n+1)4+2(n+1)3+(n+1)2+2(n+1)+1.
2n7+n6+2n4+2n3+2n2
=n7+7n6+21n5+35n4+35n3+21n2+7n+1
+n5+5n4+10n3+10n2+5n+1
+n4+4n3+6n2+4n+1
+2(n3+3n2+3n+1)
+(n2+2n+1)+2(n+1)+1.
n7-6n6-22n5-39n4-49n3-42n2-26n-9=0 .....[1].
Let A(n)=n7-6n6-22n5-39n4-49n3-42n2-26n-9.
For n>2,
A(n)<n7-6n6-22n5.
Since A(n)=0,
0<n7-6n6-22n5.
0<n5(n-3+√31)(n-3-√31).
Since n>2, n>3+√31, which means n≥9 .....[2].
Now, to differentiate A(n),
A'(n)=7n6-36n5-110n4-156n3-147n2-84n-26.
A''(n)=42n5-180n4-440n3-468n2-294n-84.
A'''(n)=210n4-720n3-1320n2-936n-294.
A(4)(n)=840n3-2160n2-2640n-936.
A(5)(n)
=2520n2-4320n-2640
=2520{(n-6/7)2-262/147}.
If A(5)(n)=0, n=6/7±√(262/147).
Therefore, A(5)(n)>0 for n≥3.
Although n≥9 according to [2], I try inspecting A(n) for n≥10.
When n≥10, A(4)(n) increases monotonously.
A(4)(10)=840000-216000-26400-936>0.
Therefore, A(4)(n)>0 for n≥10.
And when n≥10, A'''(n) increases monotonously.
A'''(10)=2100000-720000-132000-9360-294>0.
Therefore, A'''(n)>0 for n≥10.
And when n≥10, A''(n) increases monotonously.
A''(10)=4200000-1800000-440000-46800-2940-84>0.
Therefore, A''(n)>0 for n≥10.
And when n≥10, A'(n) increases monotonously.
A'(10)
=7000000-3600000-1100000-156000-14700-840-26>0.
Therefore, A'(n)>0 for n≥10.
And when n≥10, A(n) increases monotonously.
A(10)
=107-6000000-2200000-390000-49000-4200-260-9>0.
Therefore, A(n)>0 for n≥10 .....[3].
From [2] and [3], A(n)=0 only when 9≤n<10.
In consequence, n=9.
Deriving two inequalities from an equation to limit n's range
Let A=2n7+n6+2n4+2n3+2n2, and
B=(n+1)7+(n+1)5+(n+1)4+2(n+1)3+(n+1)2+2(n+1)+1.
For n>2, A<2n7+2n6, and B>(n+1)7.
From the given conditions, A=B. Then,
(n+1)7<2n7+2n6.
(n+1)6<2n6.
(n+1)6/n6<2<96/86.
(n+1)/n<9/8.
n>8.
Since 'n' is an integer, n≥9 .....[1].
On the other hand,
for n>2, A>2n7+n6, and B<(n+1)7+2(n+1)5.
From [1], 2≤(n+1)/5.
Then, B<(n+1)7+(n+1)6/5.
Since A=B,
2n7+n6<(n+1)7+(n+1)6/5.
(2n+1)n6<(n+6/5)(n+1)6.
2-7/(5n+6)<(n+1)6/n6.
From [1], 2-7/(5n+6)≥2-7/(5*9+6)=95/51.
Then, (n+1)6/n6>95/51>116/106.
(n+1)/n>11/10.
n<10 .....[2].
From [1] and [2], n=9.
Factorizing a 7th-degree polynomial of 'n'
From the given conditions,
2n7+n6+2n4+2n3+2n2
=(n+1)7+(n+1)5+(n+1)4+2(n+1)3+(n+1)2+2(n+1)+1.
(n+1)(2n6-n5+n4+n3+n2+n-1)
=(n+1){(n+1)6+(n+1)4+(n+1)3+2(n+1)2+(n+1)+2}.
(n+1)2(2n5-3n4+4n3-3n2+4n-3)
=(n+1)2{(n+1)5+(n+1)3+(n+1)2+2(n+1)+1}
=(n+1)2(n5+5n4+11n3+14n2+12n+6).
(n+1)2(n5-8n4-7n3-17n2-8n-9)=0.
(n+1)2(n2+1)(n3-8n2-8n-9)=0.
(n+1)2(n2+1)(n2+n+1)(n-9)=0.
Since 'n' is a positive integer, n=9.
Making the most of n's being an integer
From the given conditions,
2n7+n6+2n4+2n3+2n2
=(n+1)7+(n+1)5+(n+1)4+2(n+1)3+(n+1)2+2(n+1)+1 .....[1].
Using positive integers 'p' and 'q', [1] is described as:
pn=qn+9.
(p-q)n=32.
Since (p-q) is an integer and 'n' is an integer larger than 2, n=3 or n=9.
Suppose n=3,
the left side of [1] is smaller than 3*37=6561,
and the right side of [1] is larger than 47=16384.
Therefore, [1] is contradictory for n=3.
Hence, n=9.