There are three consecutive integers.nine times

I need some help here. If we take the sum of 2 integers from 1 to 9, and they can be the same number, the maximum number is 18. I don't know how you are supposed to get four consecutive digits, unless you pad the number with zeros. Am I missing something?

Good luck,

Paul

Let x be the smallest number.
Let x + 2 be the other number (consecutive even integer)

Now to translate the rest;)
three times the smaller 3(x)
19 more -19
sum of the two integers - (x) +( x+2)

Pulling it together,
3x -19 = x + x +2
collect like terms
3x - 19 = 2x + 2
Put all the constants on one side by adding 19 to both sides.
3x - 19 +19= 2x + 2 + 19
3x = 2x +21
Subtract 2x from both sides to have all the x's on one side.
3x - 2x = 2x +21 - 2x
x = 21
The other number is 21 + 2, or 23
Check:three times the smaller = 3 x 21 = 63
sum of the two integers = 21 + 23 or 44
is 63 at least 19 more than 44

38.
The three numbers are 34, 36, and 38.

Here's a simple (and inefficient) 35s program. It simply goes through a loop 25 times, adding up the numbers.

A001 LBL A
A002 25
A003 STO A
A004 0
A005 RCL+ A
A006 DSE A
A007 GTO A005

The sum of the first N positive integers can be calculated more efficiently as N(N+1)/2 .

19, 21, 23

The two numbers are 32 and 34. The rest is your homework, you do it.

X + ( X + 2 ) + ( X + 4 ) = 105
3X + 6 = 105
3X = 105 - 6
3X = 99
X = 99/3
X = 33
So 33 + 35 + 37 = 105

try this website:

http://www.homeworkspot.com/

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