Lets use the remainder theorem.

33 = 3*11
first divide any number that is divisible by 3 lets say 1053/3 = 351
so (1044*1047*1050*351)/11

1044/11 = r10
1047/11 = r2
1050/11 = r5
351/11 = r10
multiply all the r and we get 100
now 100 is greater than 11 so again divide by 11
100/11 = r10
since we divide the orginal multiplication my 3 so we have to multiply the final remainder by 3
hence 30 is the remainder and answer is C

Because you reduced you fraction by 3. That's the reason you are getting 10. Multiply 10 by 3 and you will get correct remainder 30.
If one reduce fraction by some factor to simplify the calculation then one must not forget to multiply answer by same factor in the end.
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Bunuel, Is it the correct approach?

1044*1047*1050*1053 = n*(n + 3)*(n + 6)*(n + 9)

3*6*9 = 162

162/33, remainder = 30
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What is the remainder when 30 is divided by 4?

One approach:
1. Break the dividend 30 into factors: 30 = 5*6
2. Divide the divisor 4 into each factor: 5/4 = 1 R1, 6/4 = 1 R2
3. Multiple the resulting remainders: 1*2 = 2

Step 3 indicates that 30 divided by 4 will yield a remainder of 2.
This approach can be applied to any problem that asks for the remainder when a large integer is divided by a divisor.
Repeat the 3 steps until the value yielded by Step 3 is less than the divisor.



Since 1044 has a digit sum that is a multiple of 3 (1+0+4+4=9), 1044 is divisible by 3:
1044 = 3*348
Since 1047 is 3 more than 1044, 1047 = 3*348 + 3 = 3(348+1) = 3*349
By extension:
1050 = 3*350
1053 = 3*351
Thus:
1044*1047*1050*1053 = (3*348)(3*349)(3*350)(3*351) = 81*348*349*350*351

Dividing 33 into each of the five factors in blue and multiplying the resulting remainders, we get:
15*18*19*20*21

18*20 = 360
19*21 = (20-1)(20+1) = 20²-1² = 400-1 = 399
Thus:
15*18*19*20*21 = 15*360*399

Dividing 33 into each of the 3 factors in red and multiplying the resulting remainders, we get:
15*30*3

15*30*3 = 1350
Dividing 33 into 1350, we get:
40 R30
The value in green is less than the divisor (33) and thus is the desired remainder.

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