We are asked to find an integer that is all three of these.

1. 3 more than a multiple of 9
2. 1 more than a multiple of 2, which is to say it's odd.
3. 4 more than a multiple of 5.

If we add a multiple of 9, 2 or 5, respectively to a multiple of 9, 2 or 5,
respectively, we'll get another multiple of 9, 2 or 5, respectively.  Any
multiple of all three, 9, 2, and 5 is a multiple of 90.

Therefore if we can find any number that satisfies 1,2, and 3 above, we can add
any multiple of 90 to it and get another solution.  This problem calls for a 3
digit solution, but if we can find a 1 or 2 digit solution, we can add 90 to it
one or more times and get a 3 digit number that will also satisfy 1, 2, and 3.

We look at property 1.  To get an odd number when adding 3, we must begin with
an even number

So we'll list the even multiples of 9, which are 18, 36, 54, 72, 90...

We add 3 to each and get 21, 39, 57, 75, 93 …

We look at property 3, that it is 4 more than a multiple of 5.

Every multiple of 5 ends in 0 or 5.  4 more than a number ending in 0 ends
with 4, which would make it even.  4 more than a number ending in 5 ends
with 9, which would make it odd.  We want an odd number, so it must end
in 9, and the only candidate listed above that ends with 9 is 39.  It's true
that 39 satisfies all three properties 1,2, and 3. That's becayuse

39 = 4×9+3 = 2×19+1 = 5×7+4

Trouble is, 39 only has two digits, so it doesn't qualify as a solution, but we
can add 90 to it any number of times we wish, as long as we get a 3-digit
number.  So adding 90 just once to 39 gives the three-digit number 129, and

129 = 4×18+3 = 2×64+1 = 5×25+4

We can keep getting more solutions by adding 90 over and over. There are 10
3-digit solutions to your problem:

129 = 9×14+3 = 2×64+1 = 5×25+4
219 = 9×24+3 = 2×109+1 = 5×43+4
309 = 9×34+3 = 2×154+1 = 5×61+4
399 = 9×44+3 = 2×199+1 = 5×79+4
489 = 9×54+3 = 2×244+1 = 5×97+4
579 = 9×64+3 = 2×289+1 = 5×115+4
669 = 9×74+3 = 2×334+1 = 5×133+4
759 = 9×84+3 = 2×379+1 = 5×151+4
849 = 9×94+3 = 2×424+1 = 5×169+4
939 = 9×104+3 = 2×469+1 = 5×187+4

Edwin

Let N be that unknown number.


Consider the number (N+1).


Then from the condition, (N+1) is divisible by 2; is divisible by 5; and gives the remainder of 4 when is divided by 9.


So, we should find (N+1) as the number multiple of 10, which gives the remainder of 4 when is divided by 9.


Try 10, 20, 30, 40,  and you quickly will find that  N+1 = 40 satisfies these condition.


So, N = 40-1 = 39.      ANSWER