randomperson
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- May 31, 2012
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- 3
Show that 21*18^2x+36*7^3x is divisible by 19 for all positive integers x.
Thank you!
Thank you!
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Heeeelp! This algebra/thing!
- Thread starter randomperson
- Start date
Hello, randomperson!
Are we allowed to use modulo arithmetic?
We have: .\(\displaystyle 21(18^2)^x + 36(7^3)^x\)
. . . . . \(\displaystyle =\;21(324^x) + 36(343^x)\)
. . . . . \(\displaystyle \equiv\;21(1^x) + 36(1^x)\,\pmod{19}\)
. . . . . \(\displaystyle \equiv\;21 + 36\,\pmod{19}\)
. . . . . \(\displaystyle \equiv\;57\,\pmod{19}\)
. . . . . \(\displaystyle \equiv\;0\,\pmod{19}\)
Are we allowed to use modulo arithmetic?
\(\displaystyle \text{Show that }\,21(18^{2x})+36(7^{3x})\,\text{ is divisible by 19 for all positive integers }\)
We have: .\(\displaystyle 21(18^2)^x + 36(7^3)^x\)
. . . . . \(\displaystyle =\;21(324^x) + 36(343^x)\)
. . . . . \(\displaystyle \equiv\;21(1^x) + 36(1^x)\,\pmod{19}\)
. . . . . \(\displaystyle \equiv\;21 + 36\,\pmod{19}\)
. . . . . \(\displaystyle \equiv\;57\,\pmod{19}\)
. . . . . \(\displaystyle \equiv\;0\,\pmod{19}\)
Show that 21*18^2x+36*7^3x is divisible by 19 for all positive integers x.
Thank you!
randomperson,
what you typed is equivalent to
21(18^2)x + 36(7^3)x =
21(324) + 36(343)x =
6804x + 12348x =
19152x =
19(1008x)
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If you meant what was typed in the second post,
then you must use grouping symbols, such as:
21*18^(2x) + 36*7^(3x)