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 or product is divided by a divisor.
Repeat the 3 steps until the value yielded by Step 3 is less than the divisor.



Product of the first 10 prime numbers = 2*3*5*7*11*13*17*19*23*29

Dividing 4 into each of the prime factors above yields the following remainders:
2*3*1*3*3*1*1*3*3*1 = 6*9*9

Dividing 4 into each of the factors in red yields the following remainders:
2*1*1 = 2

Since the result in blue is less than the divisor of 4, the desired remainder is 2.

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=>

2, 3, 5, 7, 11, 13, 17, 19, 23 and 29 are the first 10 prime numbers.
Of these, only 2 is an even integer. Their product is an even number, but it is not divisible by 4.
Thus, the product has remainder 2 when it is divided by 4.

Therefore, C is the answer.
Answer: C
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Hi MathRevolution

Doesnt the question stem reduce to

(2*3*5*7*11*13*17*19*23*29)/ 4

If we cancel out 2 from Numerator and denominator.
We have 2 in denominator.
So remainder should be 1.

What is wrong in this approach.Can you please help?

Posted from my mobile device

Hi chetan2u Can you please point where am I going wrong in this approach?

because 2 and 5 are factors, product units digit is 0
because 2 is the only even factor, product is not divisible by 4
thus, only option for remainder is 2
C

First 10 prime numbers are 2*3*5*7*11*13*17*19*23*29

We can see them as

2*3*x, where x= 5*7*11*13*17*19*23*29

Then it can be expressed as 6x.
The remainder of 6/4 is 2.
6x = (4x +2x)
4x/4 has a remainder of 0 and 2x/4 will always have a remainder of 2 since
a) x is not a multiple of 2 because it is a product of only prime numbers
b)the remainder cannot be three because we're dealing with integers only (x is not 1.5)
c) even if one or more of the remainders of the factors of x (5, 7, etc) was three (3*3*3...) or one, this number would be mutiplied by 2 and the remainder, again, has to be two.

Essentially we need to find remainder when 3*5*.... *29 is divided by 2. The remainder is always less than divisor itself therefore answer can be only 0 or 1. Now as there is no factor of 2 left, therefore remainder will be 1


Sent from my iPhone using GMAT Club Forum mobile app

This approach is wrong for the same reason the remainder of 10/6 (or (2*5)/(2*3)) is not equal to the remainder of 5/3.

2*3*5*7*11*13*17*19*23*29 = 6469693230 = 4*1617423307 + 2

When 2*3*5*7*11*13*17*19*23*29 = 6469693230 is divided by 4, its remainder is 2.

We can define a remainder in the following way.
If we have A = B*Q + R where 0 ≤ R < |B|, R is the remainder when A is divided by B.
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Product of first 10 prime numbers = 2*3*5*...

When divided by 4, we need remainder for this calculation: 2*3*5*... / 4

Reduce the fraction to get 3*5*... / 2

Since all other prime numbers will be odd, the product will be odd too. So upon division by 2, the remainder will be 1.

To get the remainder upon division by 4, we need to multiply the 2 back to the remainder.

Actual remainder = 1*2 = 2

Answer (C)
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15/10 Rem = 5
but when we cancel out the common factor 5,
3/2 Rem = 1

Note that the remainder depends on the divisor. If the divisor is 10, the remainder is 5 but if the divisor is 2, the remainder is 1.
If we have the remainder when the divisor is 2, we can get the remainder when the divisor is 10. We just multiply the remainder obtained back by the common factor 5. So 1*5 = 5, the remainder when 15 is divided by 10.

So in question above, you just need to multiply the 1 you obtained as remainder back by 2 to get remainder as 2.
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