Problem: When the natural numbers P and P′, with P>P′, are divided by the natural number D, the remainders are R and R′, respectively. When P′ and R′ are divided by D, the remainders are r and r′, respectively. Then:
Answer Choices:
A. r>r′ always
B. r<r′ always
C. r>r′ sometimes and r<r′ sometimes
D. r>r′ sometimes and r=r′ sometimes
E. r=r′ always
Solution:
We may write P=Q1D+R where 0<R<D, and P′=Q2D+R′ where 0≦R′<D. Therefore, PP′=(Q1D+R)(Q2D+R′)=Q1Q2D2+Q2DR+Q1DR′+RR′. But RR′= Q4D+r′. Therefore, PP′=Q1Q2D2+Q2DR+Q1DR′+Q4D+r′=D(Q1Q2D+Q2R+ Q1R′+Q4)+r′. But PP′=Q5D+r. Therefore Q5=Q1Q2D+Q2R+Q1R′+Q4 and r=r′.