+4620 Let d be a positive number such that when 109 is divided by d, the remainder is 4. Compute the sum of all possible two-digit values of d.
109 - 4 = 105
105 = 3 * 5 * 7
3 * 5 = 15
3 * 7 = 21
5 * 7 = 35
15 + 21 + 35 = 71
+295 So, let us start with the the simple equations...
\(109 mod (d) = 4\)
We know this from the remainder of four from dividing by 109.
To find the possible two digit integers that abide by this, you must take the prime factorization of the greatest integer to follow this rule, giving you
\(109 - 4 = 105\)
When you take the prime factorization of 105 from above, you get \(3*5*7\)
To find the possible two digit numbers that abide by the rules in the problem, you can get
\(3*5\)
\(5*7\)
\(7*3\)
Or just 15, 35, and 21!
When you take the sum of these, \(15+35 = 50 50+21 = 71\)
And the final answer is 71! Yay, we did it!
