Hard math questions? NEED HELP!!!?

Question

Please explain how you got the answer.
1. Find the number of integers between 500 and 700 in which the sum of the digits is 12.

2. The integers greater than 1 are arranged, 4 in each row, in 5 colomns as follows: (The X's are blanks)
2, 3, 4, 5 X
X 9, 8, 7
10,11,12,13 X
X 17,16,15,14

If the pattern is followed, in which column will the number 1001 occur?

3. The positive intergers are arranged in the pattern indicated in the diagram. What number will be found in the square for the 61st (horizontal) row and 23rd (vertical) column?

1
2,3
4,5,6
7,8,9,10
11,12,13,14,15
16,17,18,19,20,21
...

Answer 1

PROBLEM 1:

This is probably easiest done by enumerating the answers.
You start with 50? The last digit would be 7, then 51? with 6 as the last digit, etc.
507, 516, 525, 534, 543, 552, 561, 570. Now you have to skip some numbers because 58? and 59? are too much.
Next you have:
606, 615, 624, ..., 651, 660. And that is it because 67?, 68?, 69? are too much and 700 won't work either.

So the total is 8 + 7 = 15 numbers.

PROBLEM 2:

You left out a 6, but I assume you meant:
2, 3, 4, 5, X
X, 9, 8, 7, 6
10, 11, 12, 13, X
X, 17, 16, 15, 14

I think the easiest thing here is to realize that you are breaking numbers into groups of 8. So 2 through 9 have the same respective positions as 10 through 17, then 18 through 25, etc.

For each number, it is in the same position as the number that comes 8 before it and 16 before it and 24 before it, etc. In other words, if we subtract a multiple of 8, we get the same position.

Start with 1001. if you remember that 1000 is a multiple of 8 you can immediately jump to 1... If you can't remember, just subtract some known multiples of 8. 1001 subtract 800; you get 201. If you subtract 160 you get 41. If you subtract 40 you get 1. So it would be in the same position as 1. Oh wait, we don't have a 1. So instead look to 9... where is 9? Look above and you'll see It is in column 2. So there is your answer.

1001 will appear in column 2.

PROBLEM 3:

Row 1 - 1 number, ends with 1
Row 2 - 2 numbers, ends with 3
Row 3 - 3 numbers, ends with 6
Row 4 - 4 numbers, ends with 10

The last numbers are the "triangle" numbers. If you are familiar with this sequence (1, 3, 6, 10, 15, etc.) it is the sum of 1 + 2 + 3 + 4 + ... + n. And the closed formula for each row is n(n+1)/2.

In other words:
Row 1 --> (1*2) / 2 = 1
Row 2 --> (2*3) / 2 = 3
Row 3 --> (3*4) / 2 = 6
Row 4 --> (4*5) / 2 = 10
etc.
So let's jump to row 60:
Row 60 --> 60 * 61 / 2 = 1830

If the last number on row 60 is 1830, then Row 61 will look like:
1831, 1832, 1833, ..., etc., ..., 1889, 1890, 1891

To find the 23rd number, think, "What number is 23 after 1830?" It is 1853.

So the answer is 1853 will be in row 61, column 23.