Question

Two numbers, $297_B$ and $792_B$, belong to base $B$ number system. If the first number is a factor of the second number then the value of $B$ is:

• A. 11
• B. 12
• C. 15
• D. 17
• E. 19

Hint:

For base-$B$ problems, expand digits in powers of $B$. Then test given options systematically. Always double-check arithmetic for higher bases to avoid slips.

Answer 1

Answer: (E)

Step 1: Convert base-$B$ numbers to decimal.
\[ 297_B = 2B^2 + 9B + 7, \qquad 792_B = 7B^2 + 9B + 2. \] Step 2: Condition for divisibility.
We require: \[ 297_B \mid 792_B ⇒ (2B^2+9B+7) \mid (7B^2+9B+2). \] Step 3: Try answer choices.
\underline{$B=11$:}
$297_{11}=2(121)+99+7=348$, $792_{11}=847$. $847/348$ is not integer.
\underline{$B=12$:}
$297_{12}=2(144)+108+7=403$, $792_{12}=1186$. $1186/403$ not integer.
\underline{$B=15$:}
$297_{15}=2(225)+135+7=592$, $792_{15}=1612$. $1612/592$ not integer.
\underline{$B=17$:}
$297_{17}=2(289)+153+7=738$, $792_{17}=2080$. $2080/738$ not integer.
\underline{$B=19$:}
$297_{19}=2(361)+171+7=900$, $792_{19}=2542$. $2542/900$ not integer. Wait check carefully. Step 4: Re-check calculation for $B=19$.
\[ 297_{19}=2(361)+9(19)+7=722+171+7=900, \] \[ 792_{19}=7(361)+9(19)+2=2527+171+2=2700. \] Now $2700/900=3$, an integer. \[ \boxed{19} \]