Show that if the remainder of c divided by d is 1, then the same is true for the remainder of c^2.
2006-10-24 02:43:55 · 4 answers · asked by Psychic K 1 in Science & Mathematics ➔ Mathematics
Let c = n.d + 1, then c^2 = n^2.d^2 + 2n.d + 1 = m.d + 1. You can pick out the value of m in terms of n and d, but you don't need to - all you need is for m to be an integer, and it is.
2006-10-24 02:51:01 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Lets assume C is 4 and D is 3.
then C/D = 4/3= 1; remainder=1
Also D/2 = 3/2=1;remainder=1
2006-10-24 09:49:42 · answer #2 · answered by MY Regards to All 4 · 0⤊ 0⤋
There exists an integer m such that c = m*d + 1. Therefore, c^2 = m^2 * d^2 + 2*m*d + 1 => c^2 = (m^2 * d + 2m)d + 1. Put k = m^2 *d + 2m. Then, k is an integer, and we conclude c^2 = k*d + 1 for some integer k. This shows the remainder of c^2divided by d is 1.
2006-10-24 10:17:55 · answer #3 · answered by Steiner 7 · 0⤊ 0⤋
c=dq+1
c^2=(dq+1)^2=d^2q^2+2dq+1
c^2=dq(dq+2)+1
q(dq+2)=n so
c^2=dn+1
2006-10-24 09:54:41 · answer #4 · answered by farsh m 1 · 0⤊ 0⤋