Let n and m be integer numbers. Prove that, if 3n+7m
is divisible by 19, then the number 43n + 75m is divisible by 19 also.
If 3n+7m is 0 mod 19, then m = mod(5n,19).
(from observing a spreadsheet table)
Verify this:
Call x = mod((3n+7m),19) and substitute m = mod(5n,19).
x = mod((3n+7(mod(5n,19))),19)
x = mod((3n+mod(35n,19)),19)
x = mod(mod(38n,19),19) = 0
Starting with 43n + 75m, this equals 43n + 75*mod(5n,19).
= 43n + mod(375n,19).
The mod value of this is the same as mod((375n+43n),19)
= mod(418n, 19).
But 418 is 19*22, so the expression is 0 mod 19
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Posted by Larry
on 2025-01-31 10:40:48 |
solution
