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What is the remainder when 1044*1047*1050*1053 is

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by GMATGuruNY » Mon Jul 02, 2018 1:46 am
M7MBA wrote:What is the remainder when 1044*1047*1050*1053 is divided by 33?

A. 3
B. 27
C. 30
D. 21
E. 18
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by Shahrukh_mbabreakspace » Mon Jul 02, 2018 2:23 am
We will have to find remainder for individual number:
For 1044, remainder is 21 or -12
For 1047 , remainder is 24 or -9 (just add 3)
For 1050 , remainder is 27 or -6 (just add 3)
For 1053 , remainder is 30 or -3 (just add 3)

Now, multiply the remainders (I prefer negative ones here as they are smaller in terms of magnitude)
We get 1944, now divide it by 33, we get remainder as 30
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by Vincen » Mon Jul 02, 2018 2:55 am
M7MBA wrote:What is the remainder when 1044*1047*1050*1053 is divided by 33?

A. 3
B. 27
C. 30
D. 21
E. 18

The OA is C.

Is there a fast way to solve this PS question? I'd really appreciate any help here. Thanks.
Hello M7MBA.

This is how I'd solve it.

First, note that all the numbers are divisible by 3. Now, we have $$\frac{1044\cdot1047\cdot1050\cdot1053}{3}=\frac{1044}{3}\cdot1047\cdot1050\cdot1053$$ $$=348\cdot1047\cdot1050\cdot1053$$ Now, we can divide each number by 11 and we get:

Nº -------- Remainder
348 ------- 7
1047 ----- 2
1050 ----- 5
1053 ----- 8

Now, the 7*2*5*8 = 560 and the remainder when 560 is divided by 11 is equal to 10.

Finally, since we divided by 3 at the beginning, here we revert back multiplying by 3, so we get that the remainder is 10*3=30.

Therefore, the correct answer is the option [spoiler]C=30[/spoiler].

I hope it helps you. <i class="em em-smiley"></i>
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by Scott@TargetTestPrep » Mon Apr 29, 2019 8:08 am
M7MBA wrote:What is the remainder when 1044*1047*1050*1053 is divided by 33?

A. 3
B. 27
C. 30
D. 21
E. 18

The OA is C.

Is there a fast way to solve this PS question? I'd really appreciate any help here. Thanks.
We need to use the following fact:

Let R(n, d) denote the remainder when positive integer n is divided by positive integer d. Then R(ab, c) = R(R(a, c) * R(b, c), c).
For example, let a = 12, b = 8, and c = 5, we have R(12 * 8, 5) = R(96, 5) = 1, R(12, 5) = 2, R(8, 5) = 3, and R(2 * 3, 5) = 1. We see that R(12 * 8, 5) = R(R(12, 5) * R(8, 5), 5).

So, instead of dividing the product by 33, let's divide each individual factor by 33 and determine its remainder.

Since 1044 = 33 * 31 + 21, R(1044, 33) = 21 and thus R(1047, 33) = 24, R(1050, 33) = 27 and R(1053, 33) = 30.

Now, we can find the remainder when 21 * 24 * 27 * 30 is divided by 33. Although this is much better than the original product, it's still quite a hassle. However, notice that

21 * 24 * 27 * 30 = (33 - 12)(33 - 9)(33 - 6)(33 - 3)

When we expand the product on the right hand side using FOIL, every term will have a factor of 33, except the last term which is 12 * 9 * 6 * 3, which is much easier to handle than 21 * 24 * 27 * 30.

R(12 * 3, 33) = R(36, 33) = 3 and R(9 * 6, 33) = R(54, 33) = 21, and finally,

R(3 * 21, 33) = R(63, 33) = 30

So the remainder must be 30.

Answer: C

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by GMATGuruNY » Tue Apr 30, 2019 2:50 am
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.
M7MBA wrote:What is the remainder when 1044*1047*1050*1053 is divided by 33?

A. 3
B. 27
C. 30
D. 21
E. 18
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.

The correct answer is C.
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