# Two algebra problems? Yes I am a little stupid and even if you can do one, that's ok!

• Jun 1st 2012, 04:17 AM
someoneanonymous
Two algebra problems? Yes I am a little stupid and even if you can do one, that's ok!
1. Show that 21*18^(2x)+36*7^(3x) is divisible by 19 for all positive integers x

2. Find all pairs of odd integers m and n which satisfy the following equation:
m+128n=3mn
• Jun 1st 2012, 06:52 AM
Soroban
Re: Two algebra problems? Yes I am a little stupid and even if you can do one, that's
Hello, someoneanonymous!

Quote:

$\text{1. Show that }\,N \:=\:21(18^{2x})+36(7^{3x})\,\text{ is divisible by 19 for all positive integers }x.$

We have: . $21(18^2)^x + 36(7^3)^x$

. . . . . . $=\;21(324)^x + 36(343)^x$

. . . . . . $\equiv\;21(1)^x + 36(1)^x \pmod{19}$

. . . . . . $\equiv\;21 + 36 \pmod{19}$

. . . . . . $\equiv\;57 \pmod{19}$

. . . . . . $\equiv\;0 \pmod{19}$

Therefore, $N$ is divisible by 19.