Perfect Squares with CongruencesDate: 02/16/2002 at 14:17:15 From: Stacey Subject: Perfect squares with congruences Prove that there is no perfect square a^2 whose last digits are 35. Date: 02/17/2002 at 17:42:47 From: Doctor Paul Subject: Re: Perfect squares with congruences What if I asked you to prove that there is no perfect square a^2 whose last digit is 7? Notice that every integer falls into one of 10 classes: those that are multiples of 10 1 more than a multiple of ten 2 more than a multiple of ten 3 more than a multiple of ten 4 more than a multiple of ten 5 more than a multiple of ten 6 more than a multiple of ten 7 more than a multiple of ten 8 more than a multiple of ten 9 more than a multiple of ten Written in terms of moduli, we can say that every integer is either congruent to 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 mod 10. Notice that the last digit of a^2 is just the same thing as computing a^2 mod 10. So we do this for the 10 possibilities: If an integer is congruent to 0 mod 10, then x^2 = 0^2 = 0 mod 10 i.e., if x = 0 mod 10 then the last digit of x^2 is 0. If an integer is congruent to 1 mod 10, then x^2 = 1^2 = 1 mod 10 i.e., if x = 1 mod 10 then the last digit of x^2 is 1. If an integer is congruent to 2 mod 10, then x^2 = 2^2 = 4 mod 10 i.e., if x = 2 mod 10 then the last digit of x^2 is 4. If an integer is congruent to 3 mod 10, then x^2 = 3^2 = 9 mod 10 i.e., if x = 3 mod 10 then the last digit of x^2 is 9. If an integer is congruent to 4 mod 10, then x^2 = 4^2 = 16 = 6 mod 10 i.e., if x = 4 mod 10 then the last digit of x^2 is 6. . . [fill in the details] . If an integer is congruent to 9 mod 10, then x^2 = 9^2 = 81 = 1 mod 10 i.e., if x = 9 mod 10 then the last digit of x^2 is 1. This shows that no square has 7 as its last digit. A similar argument holds for showing that the last two digits of a number will never be 35. Just square each of the 100 residue classes mod 100 and show that none of them yields 35. It's computationally time-consuming so I wrote a little Maple script to do it for me: > restart; > L := [seq(x[i],i=1..100)]: > for i from 1 to 100 do > L[i]:=i^2 mod 100; > od: > L; [1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 21, 44, 69, 96, 25, 56, 89, 24, 61, 0, 41, 84, 29, 76, 25, 76, 29, 84, 41, 0, 61, 24, 89, 56, 25, 96, 69, 44, 21, 0, 81, 64, 49, 36, 25, 16, 9, 4, 1, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 21, 44, 69, 96, 25, 56, 89, 24, 61, 0, 41, 84, 29, 76, 25, 76, 29, 84, 41, 0, 61, 24, 89, 56, 25, 96, 69, 44, 21, 0, 81, 64, 49, 36, 25, 16, 9, 4, 1, 0 ] Inasmuch as none of the squares mod 100 is 35, we can conclude that there is no number whose square has 35 as its last two digits. I hope this helps. Please write back if you'd like to talk about this more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
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