<p>All numbers divisible by both 4 and 15 are also divisible by which of the following?</p>
<p>(A) 6
(B) 8
(C) 18
(D) 24
(E) 45</p>
<p>Well the least common multiple of 4 and 15 is 60. Of the choices, only 6 divides into 60 evenly. That wasn’t algebraically, but I hope it helped!</p>
<p>Thanks! That helped!</p>
<p>There is a slightly more general way of solving problems such as this.</p>
<p>All numbers divisible by both M and N are also divisible by which numbers?</p>
<p>Rewrite M as the product of its prime factors, i.e. m1<em>m2</em>m3 etc.
Rewrite N as the product of its prime factors, i.e. n1<em>n2</em>n3 etc.</p>
<p>So any number divisible by both M and N is also divisible by any combination of factors of M AND factors of N. If the prime factor appears in both M and N one of the options is the common factor.</p>
<p>The prime factorization of M is 4=2<em>2, and of N 15=3</em>5. So the answers include 2<em>3, 2</em>5, etc.</p>
<p>Try the same with M=27 and N=51. Here the prime factorization of M is 27=3<em>3</em>3, and of N=3*17. So the answers include 3, 51, 153 etc. 3 is a possible choice because it is a common factor of M and N.</p>
<p>This is a consequence of a basic theorem in arithmetic: integers have unique factorization into primes.</p>