Originally posted by David113true...
True or false:
If n is an integer, and n squared minus 5 is divisible by an integer m, then the last digit of m cannot be 3 or 7.
for it to work, the last digit of the squared number (after being squared) would have to be a 2 or an 8
meaning that n must be even (even number times even number equals even number and viceversa for odd numbers)when two numbers are squared, the last digit of the product is determined by the last digit of n, which must be 2 4 6 8 or 0
these squares are 4, 16, 36, 64, and 0
since we are concerned about the last digit of these numbers, we can isolate them as:
4, 6, 6, 4, and 0
meaning that the square of any number will never have a 2 or 8 as its last digit...
so the last digit of m will never be 3 or 7 after subtracting 5...
solved.
Originally posted by XanthosNZi answered his original question and provided an explanation on top...
Looks like you can't read.
i was correct in my explanation...
so i dont see what u mean i cant read...
it is a true statement that n squared minus 5 never holds a 3 or a seven as the final digit as long as n meets the definition of an integer...
puzzle solved, we can go home now