Problem:
If the positive integer c has positive integer divisors a and b with c=ab, then a and b are said to be complementary divisors of c. Suppose that N is a positive integer that has one complementary pair of divisors that differ by 20 and another pair of complementary divisors that differ by 23 . What is the sum of the digits of N?
Answer Choices:
A. 9
B. 13
C. 15
D. 17
E. 19
Solution:
Write the factorizations as a(a+20)=b(b+23)=N. Thus 0<b<a. Multiply through by 4 and complete the square to see that
(2a+20)2=(2b+23)2−129.
Thus
(2b+23)2−(2a+20)2=(2b+2a+43)(2b−2a+3)=129=3⋅43.
The positive divisors of 129 are 1,3,43, and 129 . The number 2b+2a+43 is a divisor of 129 that is greater than 43 , so
2b+2a+43=129
and
2b−2a+3=1
This gives a=22 and b=21, so a+20=42,b+23=44, and hence N=22⋅42=22⋅3⋅7⋅11= 21⋅44=924. The sum of its digits is 9+2+4=(C)15​.
OR
As in the previous solution, if the factorizations are written as a(a+20)=b(b+23)=N, then b<a and a+20<b+23, from which b<a<b+3.
Because a and b must be integers, a must equal b+1 or b+2. In the case when a=b+1,a(a+20) becomes (b+1)(b+21)=b2+22b+21. Setting this equal to b(b+23)=b2+23b yields the solution b=21, which corresponds to N=21⋅44=22⋅42=924, and the sum of the digits of 924 is (C)15​.
In the case when a=b+2,a(a+20) becomes (b+2)(b+22)=b2+24b+44. Setting this equal to b(b+23)=b2+23b yields the only solution, b=−44. This does not meet the requirement that the complementary divisors be positive, although here as well (−44)⋅(−21)=(−42)⋅(−21)=924.
The problems on this page are the property of the MAA's American Mathematics Competitions