Let N = 123456789101112...4344 be the 79-digit number that is formed

i reckon the end of
- 45 x odd ends with 5
- 45 x even ends with 0
Thus the remainder will either be
4-5 = ends with 9 OR (Option C)
4-0 = ends with 4 (Option B/E) but i cant find a pattern to determine this further - can someone help explain this please! thanks a lot

I won't discuss whether this type of question will be tested on the GMAT or not; I though believe such concept won't be tested.
We will break 45 into two co prime factors 9 and 5.
With 5 as divisor, we get 4 as the remainder as the last digit of the 79 digit number is 4.
To find remainder when the number is divided by 9, we can simply add the numbers from 1 to 44.
1+2+3+.....+44=(44*45)/2=22*45. This is clearly divisible by 9 giving 0 as the remainder.

With 5 remainder is 4 or 9 or 14 or 19 etc.
With 9 remainder is 0 or 9 or 18 or 27 etc.

Therefore, Remainder of the no when divide by 45 will be 9

My two cents (How I solved this). This is a super-fun question!

N = 1234....4344 (number formed by writing integers from 1 to 44 in order).

To find: Remainder when N is divided by 45.

For a number to be divisible by 45, it should be divisible by both 5 and 9.

Point: If we can find the closest number to N which is divisible by 5 (ends with 5 or 0) and also divisible by 9 (sum of all digits = 9), from there, we can easily find the remainder when N is divided by 45.

(1) Let's check if N is divisible by 9.

Sum of digits = sum of unit digits + sum of tens digits

Sum of units digits = Sum of units digits from 1 to 9 + sum of units digits from 10 to 19 + sum of units digits from 20 to 29 +.... sum of units digits from 40 to 44.

= [1+..9] + [0+1+...9] + [0+1+...9] + [0+1+...9] + [0+1+2+3+4]

Sum of digits from 0 to 9 = 45.

Sum of unit digits = 4x45 + 10 = 180+10 = 190.

Sum of tens digits = sum of tens digits of 10 to 19 + sum of tens digits of 20 to 29 + sum of tens digits of 30 to 39 + sum of tens digits of 40 to 44

Sum of tens digits = 10(1) + 10(2) + 10(3) + 5(4) = 80.

Therefore,
Sum of digits of N = 190 + 80 = 270 = multiple of 9.

Therefore, N is divisible by 9.

Now, close to N, if we can find a number that is divisible by 9 but also by 5 (ending with 5 or 0), then, that number is divisible by 45.

Observe:

1234....4335, 1234........4344, 1234......4353, 1234.........4362, 1234.......4371, 1234..........80.

We can use either of the two numbers highlighted above.

1234......4335 is divisible by 45 because it is divisible by both 9 and 5. Our N is 9 more than this number. Hence, the remainder when N is divided by 45 should be 9.

Alternately, 1234....4380 is divisible by 45. Our N is 36 less than this number. Hence, the remainder when N is divided by 45 should be -36 i.e., +(45-36) = 9.

Choice C.

Hope this helps!
Harsha