Expert: Sherry Wallin - 4/18/2009

Question
2. how many positive integers less than 1 000 000 have the sum of their digits equal to 19?


3. Let X = {1, 2, 3, ..., 1000}. Find the number of 3-element subsets {a, b, c} of X such that the product abc is divisible by 3.

5. Suppose a particle is located at the origin of the xy-plane. In how many ways can the particle move from the origin to the point (5,4) if the only allowable moves are one unit up or one unit to the right and it must avoid point (3,2)?

pls dont get it busy ... i need solution plsss/... plss help me... thnx

Answer
Carl~
    A couple of quick thoughts about #2. How many ways can you get a sum of 19 with 3 digits? 1+9+9 = 19 = 2+8+9 = 3+7+9 = 4+6+9 = 5+5+9
8+8+3 = 7+7+5 = 6+6+7 = 5+6+8 = ... there may be more, find them, and then figure out all the permutations of each. For example the digits (2,8,9), (9,2,8), (8,9,2), (2,9,8), (9,8,2), (8,2,9) are 6 different arrangements, why? Because each digit can take on 3 different places so this is a counting problem 3*2*1 = 6 ways to arrange the 3 digits 2,8,9
Figure this out for all the different ways to sums to 19 and then add the sums of the permutations together.


Send me the questions again if you want more answers and explanations.

Math Prof

Ok I have found 10 different ways to find the sum of 19:
1+9+9, 2+8+9, 3+8+8, 3+7+9, 4+7+8, 4+6+9, 5+7+7, 5+8+6, 5+5+9, 6+6+7. Now we need to count the number of arrangements: notice when there are duplicate numbers as in 1+9+9 there are only 3 different arrangements:
1+9+9, 9+1+9, 9+9+1 so we have 5 sums with double digits and 4 without so we have 5*3 arrangements and 5*6 arrangements for a total of 15+30 = 45 arrangements. Now the interesting part comes in because now you have to figure out how many numbers between 1 and 1 million have those digits only. It is not as hard as it seems though because all other digits have to be 0. For 1 to 1000 it is just 45 ways, now count from 1001 thru 9999. For 4 digits numbers only 3 can be a non zero digit and the other 1 has to be a zero. So you have each of
1+9+9, 2+8+9, 3+8+8, 3+7+9, 4+7+8, 4+6+9, 5+7+7, 5+8+6, 5+5+9, 6+6+7 with a zero in them in each place or 4*3*2*1 = 24 ways to arrange 4 different digits. So there are 5 arrangements with no duplicate digits and 5 arrangements with a set of duplicate digits. 5*24 + 5*12
=120+60 = 180 4 digit ways for a sum of 3 digits to equal 19.
So now we have 45 + 180 = 225 integers between 1 and 9,999 with a sum of 19 in it made with 3 digits. Now it remains to be seen ho many 5 digit numbers with 3 nonzero digits that sum to 19 that are in numbers between 10,000 and 99,999 and from 100000 thru 999,999. Using the same process we proceed counting the number of ways with 5 digits
5*4*3*2*1 = 120 and for 5 of those sums there are going to be 120 ways to arrange the digits and half that many to arrange those digits that are duplicated. 5*120 + 5*60 = 600 + 300 = 900. Add 900 to our previous sum (225 + 900 = 1125) and count from 100000 thru 999,999. There are 6! =6*5*483*2*1 = 720 ways to arrange 3 digits that sum to 19 in a 6 digit number. So 5*720 + 5*360 = 3600 + 1800 = 5400. Now add 5400 to 1125 getting 6525 numbers between 1 and 1 million with digits that sum to 19.