Find 2 consecutive even integers such that the smaller added to four ties the lager gives a sum of 48

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

129

If this is homework, be sure to show your work.

83+85+87+89=344

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 .

Let x be first (smaller) integer and y second (larger) integer.

1. They are consecutive (and even) : y=x+2
2. Twice the smaller exceeds the larger by 18: 2*x=y+18
3. Substitute first equation into second: 2x=x+2+18
4. 2x-x=2+18
5. x=20
6. y=x+2=22

These two numbers are 20 and 22.

We'll call x the smaller integer and y the larger. So we know x+3y=22. We can rearrange this to say:
x=22-3y. Now we guess, I'll guess y=6, now we solve and get x=22-3(6)=4. 4 and 6 are two even consecutive integers! So the answer is 4 and 6.

The sum of two consecutive even integersis Even.