# Determine the smallest value

• Oct 5th 2012, 11:42 AM
Lukaszm
Determine the smallest value
Troubles again (Giggle)
Determine the smallest value of $\displaystyle |20^m-9^n|$ ; m and n are positive integers.
I created a graph of it and as far as I've noticed 11 is the smallest value.
The thing is- how to prove it?
I tried to play with divisibility by 11, but I didn't succeed.
Any ideas?
Regards,
Lukasz
• Oct 5th 2012, 02:04 PM
TheEmptySet
Re: Determine the smallest value
Quote:

Originally Posted by Lukaszm
Troubles again (Giggle)
Determine the smallest value of $\displaystyle |20^m-9^n|$ ; m and n are positive integers.
I created a graph of it and as far as I've noticed 11 is the smallest value.
The thing is- how to prove it?
I tried to play with divisibility by 11, but I didn't succeed.
Any ideas?
Regards,
Lukasz

we can write twenty as

$\displaystyle 20^m=(9+11)^m$

If we expand using the binomial theorem we get

$\displaystyle (9+11)^m=9^m+\binom{m}{1}9^{m-1}\cdot 11+ \cdots +\binom{m}{m-1}9\cdot11^{m-1}+11^{m}=\sum_{i=0}^{m}\binom{m}{i}9^{m-i}11^i$

Now if we subtract we get

$\displaystyle \bigg| 9^m-9^n+\sum_{i=1}^{m}\binom{m}{i}9^{m-i}11^i\bigg|$

If we set m=n the first two terms reduce out.

Now since all terms of the sum have the same sign we must minimize the sum. the value m=1 gives this minimum

This should give you the conclustion that you are looking for.
• Oct 5th 2012, 02:50 PM
johnsomeone
Re: Determine the smallest value
Here's a proof that k = 11.

Let $\displaystyle k = 20^m - 9^n$ s.t. $\displaystyle |k|$ is a min among such numbers, $\displaystyle n, m \in \mathbb{Z}^+$.

Look at it modulo 10: $\displaystyle k \equiv (0) - (-1)^n \equiv (-1)^{n+1} \mod 10$.

Thus $\displaystyle |k| \in \{ 1, 9, 11, 19, 21, 29, 31, ...\}$.

Since $\displaystyle 11 = 20^1 - 9^1$, and $\displaystyle |k|$ is defined to be minimal, conclude that $\displaystyle |k| \in \{ 1, 9, 11 \}$.

But 9 can be excluded, for if $\displaystyle |k| = 9$, then $\displaystyle 20^m - 9^n = \pm 9$, which is impossible (look $\displaystyle \mod 9$, recall $\displaystyle n >0$).

Thus $\displaystyle |k| \in \{ 1, 11 \}$.

Also, you can exclude $\displaystyle k = 1$, because:

Assume $\displaystyle k=1$. Then $\displaystyle 20^m - 9^n = 1$, which implies $\displaystyle 20^m - 1 = 9^n$.

But 19 divides $\displaystyle 20^m - 1$, so 19 divides $\displaystyle 9^n$, a contradiction. Thus $\displaystyle k \ne 1$.

Thus $\displaystyle k \in \{ -1, 11 \}$.

Need to show that $\displaystyle 20^m = 9^n - 1$ has no solutions, and then 11 will be the answer.

Look mod 10 again, and see that n must be even. Thus need only show that $\displaystyle 20^m = 81^n - 1$ has no solutions, and then 11 will be the answer.

Look at $\displaystyle 20^m = 81^n - 1$ using mod 19: $\displaystyle 1 \equiv (5)^n - 1\mod 19$, so $\displaystyle (5)^n \equiv 2 \mod 19$

Now $\displaystyle \mod 19$, have: $\displaystyle 5^2 \equiv 6, 5^4 \equiv 36 \equiv -2, 5^8 \equiv (-2)^2 \equiv 4, 5^9 \equiv 20 \equiv 1$.

Filling in: $\displaystyle 5^3 \equiv 30 \equiv 11, 5^5 \equiv (-2)(5) \equiv -10 \equiv 9, 5^6 \equiv 5^45^2 \equiv -12 \equiv 7,$

and $\displaystyle 5^7 \equiv 35 \equiv -3 \equiv 16$.

Therefore $\displaystyle (5)^n \equiv 2 \mod 19$ has no solutions.

Therefore $\displaystyle 20^m = 81^n - 1$ has no solutions.

Therefore $\displaystyle 20^m = 9^n - 1$ has no solutions.

Therefore, k = 11.