Question

If \(N\) is the largest integer that will divide \(1305\), \(4665\), and \(6905\) leaving the same remainder, then the sum of all prime numbers that divide \(N\) is:

• A. \(10\)
• B. \(14\)
• C. \(17\)
• D. \(19\)

Hint:

For "same remainder" problems, immediately think of the GCD of pairwise differences.

Answer 1

The Correct Option is B

Concept: When several numbers leave the same remainder upon division by \(N\), then \(N\) divides all pairwise differences of those numbers. Thus, \[ N=\gcd(4665-1305,\;6905-4665,\;6905-1305) \]

Step 1: Find the differences. \[ 4665-1305=3360 \] \[ 6905-4665=2240 \] \[ 6905-1305=5600 \]

Step 2: Find the greatest common divisor. \[ \gcd(3360,2240)=1120 \] \[ \gcd(1120,5600)=1120 \] Hence \[ N=1120 \]

Step 3: Prime factorization of \(1120\). \[ 1120=112\times10 \] \[ =(2^4\times7)\times(2\times5) \] \[ =2^5\times5\times7 \] Prime divisors are \[ 2,\;5,\;7 \]

Step 4: Find their sum. \[ 2+5+7=14 \] Therefore, \[ \boxed{14} \]