Let be #N# the smallest integer with 378 divisors. If #N = 2^a xx 3^b xx 5^c xx 7^d#, what is the value of #{a,b,c,d} in NN# ?

1 Answer
Oct 4, 2016

Answer:

#(a, b, c, d) = (6, 5, 2, 2)#

#N = 2^6xx3^5xx5^2xx7^2=19,051,200#

Explanation:

Given a number #n# with prime factorization #n = p_1^(alpha_1)p_2^(alpha_2)...p_k^(alpha_k)#, each divisor of #n# is of the form #p_1^(beta_1)p_2^(beta_2)...p_k^(beta_k)# where #beta_i in {0, 1, ..., alpha_i}#. As there are #alpha_i+1# choices for each #beta_i#, the number of divisors of #n# is given by
#(alpha_1+1)(alpha_2+1)...(alpha_k+1)=prod_(i=1)^k(alpha_i+1)#

As #N=2^axx3^bxx5^cxx7^d#, the number of divisors of #N# is given by #(a+1)(b+1)(c+1)(d+1) = 378#. Thus, our goal is to find #(a, b, c, d)# such that the above product holds and #2^axx3^bxx5^cxx7^d# is minimal. As we are minimizing, we will assume from this point onward that #a>=b>=c>=d# (if this were not the case, we could swap exponents to get a lesser result with the same number of divisors).

Noting that #378 = 2xx3^3xx7#, we can consider the possible cases in which #378# is written as a product of four integers #k_1, k_2, k_3, k_4#. We can inspect these to see which produces the least result for #N#.

Format: #(k_1, k_2, k_3, k_4) => (a, b, c, d) => 2^axx3^bxx5^cxx7^d#


#(2, 3, 3^2, 7) => (8, 6, 2, 1) => ~3.3xx10^7#
#(2, 3, 3, 3*7) => (20, 2, 2, 1) => ~1.7xx10^9#
#color(red)((3, 3, 2*3, 7) => (6, 5, 2, 2) => ~1.9xx10^7)#
#(3, 3, 3, 2*7) => (13, 2, 2, 2) => ~9.0xx10^7#
#(1, 3, 2*3^2, 7) => (17, 6, 2, 0) => ~2.4xx10^9#

We can stop here, as any further cases will have some #k_i>=27#, giving #2^a >= 2^26 ~~ 6.7xx10^7#, which is already greater than our best case.

By the above work, then, the #(a, b, c, d)# which produces a minimal #N# with #378# divisors is #(a, b, c, d) = (6, 5, 2, 2)#, giving #N = 2^6xx3^5xx5^2xx7^2=19,051,200#