Question

If $a$, $b$ and $c$ are positive integers less then 10, then $(10a+b)(10a+c)$ equals $100a(a+1)+bc$ if;

• A. $b+c=10$
• B. $b=c$
• C. $a+b=10$
• D. $a=b$
• E. $a+b+c=10$

Answer 1

The correct option is A $b+c=10$

$(10a+b)(10a+c)=100a^2+10a(b+c)+bcGiven:(10a+b)(10a+c)=100a(a+1)+bc\therefore 100a^2+10a(b+c)+bc=100a(a+1)+bc⟹10a(b+c)=100a\therefore b+c=10$