MathDB

20

Part of 2017 AMC 10

Problems(2)

Digital Sum Returns

Source: 2017 AMC10A #20, 2017 AMC12A #18

2/8/2017
Let S(n)S(n) equal the sum of the digits of positive integer nn. For example, S(1507)=13S(1507) = 13. For a particular positive integer nn, S(n)=1274S(n) = 1274. Which of the following could be the value of S(n+1)S(n+1)?
<spanclass=′latex−bold′>(A)</span> 1<spanclass=′latex−bold′>(B)</span> 3<spanclass=′latex−bold′>(C)</span> 12<spanclass=′latex−bold′>(D)</span> 1239<spanclass=′latex−bold′>(E)</span> 1265<span class='latex-bold'>(A)</span>\ 1 \qquad<span class='latex-bold'>(B)</span>\ 3\qquad<span class='latex-bold'>(C)</span>\ 12\qquad<span class='latex-bold'>(D)</span>\ 1239\qquad<span class='latex-bold'>(E)</span>\ 1265
AMCAMC 10AMC 10 AAMC 12AMC 12 A2017 AMC 10A2017 AMC 12A
Extraneous Information

Source: 2017 AMC 10B #20 / AMC 12B #16

2/16/2017
The number 21!=51,090,942,171,709,440,00021!=51,090,942,171,709,440,000 has over 60,00060,000 positive integer divisors. One of them is chosen at random. What is the probability that it is odd?
<spanclass=′latex−bold′>(A)</span>121<spanclass=′latex−bold′>(B)</span>119<spanclass=′latex−bold′>(C)</span>118<spanclass=′latex−bold′>(D)</span>12<spanclass=′latex−bold′>(E)</span>1121<span class='latex-bold'>(A)</span> \frac{1}{21} \qquad <span class='latex-bold'>(B)</span> \frac{1}{19} \qquad <span class='latex-bold'>(C)</span> \frac{1}{18} \qquad <span class='latex-bold'>(D)</span> \frac{1}{2} \qquad <span class='latex-bold'>(E)</span> \frac{11}{21}
2017 AMC 10BAMC 10AMCprobability2017 AMC 12BAMC 12