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Solution



Given
In this question, we are given that
    • The number x = 1998 * 1999 * 2000

To find
We need to determine
    • The remainder when x is divided by 7

Approach and Working out

1998 = 7 * p + 3
1999 = 7 * p + 4
2000 = 7 * p + 5, where p is a positive integer.

    • Hence, the individual remainders are 3, 4, and 5 respectively
    • So, remainder when x is divided by 7 = remainder when 3 * 4 * 5 is divided by 7 = remainder when 60 is divided by 7 = 4

Thus, option E is the correct answer.

Correct Answer: Option E
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Solution



Given
In this question, we are given that
    • The number x = 1998 * 1999 * 2000

To find
We need to determine
    • The remainder when x is divided by 7

Approach and Working out

1998 = 7 * p + 3
1999 = 7 * p + 4
2000 = 7 * p + 5, where p is a positive integer.

    • Hence, the individual remainders are 3, 4, and 5 respectively
    • So, remainder when x is divided by 7 = remainder when 3 * 4 * 5 is divided by 7 = remainder when 60 is divided by 7 = 4

Thus, option E is the correct answer.

I understood everything but how did you know that 1998 = 7 * p + 3?

Correct Answer: Option E
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x=1998∗1999∗2000
x=(2000-2)*(2000-1)*2000
Remainder of 2000/7 => 5
Remainder of 2/7 => 2
Remainder of 1/7 => 1

x=(5-2)*(5-1)*5
Since 60 > 7 , x=60/7
Remainder of 60/7 => 4

E
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Since none of the numbers are divisible, we find the individual remainders for each number.

Therefore R[1998/7] * R[1999/7] * R[2000/7] = 3 * 4 * 5 = 60

Since the product of the remainders i.e 60, is greater than the divisor i.e 7, we redivide.
Therefore R[60/7] = 4

Option E

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@CrackVerbalGMAT- Dear Arun Sir- I liked your solution but I have one doubt. What if we had like 10 such numbers in the Q. Will we divide each number by 7 and then find the remainder? Can we have a shortcut here pls? Thanks in advance.
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@CrackVerbalGMAT- Dear Arun Sir- I liked your solution but I have one doubt. What if we had like 10 such numbers in the Q. Will we divide each number by 7 and then find the remainder? Can we have a shortcut here pls? Thanks in advance.

Hello Gauri

The method above is a short cut. In fact if all 10 numbers are not in succession, that is they do not follow a progression then the above method is the best.

Suppose you have 5 consecutive numbers, then just find the remainder for the first number and the remainders for the successive numbers will be one larger than the previous one. You then do not have to divide each number to find the remainder.

For e.g Remainder of \(\frac{113 * 114 * 115 * 116 * 117}{11}\)

Since the numbers are successive, then we need to find the remainder of \(\frac{113}{11}\) = 3

Then the remainders of the next 4 numbers are 4, 5, 6 and 7

Therefore \(R[\frac{113 * 114 * 115 * 116 * 117}{11}]\) = \(\frac{3 * 4 * 5 * 6 * 7}{11}\)

We then multiply these values and see if the product is > 11. If it is then redivide (Remainder has to be less than the divisor)

= \(R[\frac{2520}{11}]\) = 1



If you feel that multiplication of these numbers are cumbersome, then break them into parts in any order.


Find the remainders of \(R[\frac{3 * 4 * 5 }{11}]\) * \(R[\frac{6 * 7}{11}]\) = \(R[\frac{60}{11}]\) * \(R[\frac{42}{11}]\)

For the first part, we get the remainder as 5 and the second part we get the remainder as 9

Multiply these 2 to see if the product is > 11. If yes then redivide. \(R[\frac{5*9}{11}]\) = 1

Hope this Helps

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@CrackVerbalGMAT- Arun Sir- Very clear. You have told a trick which will save immense time. Never knew we can solve remaiders this way. I used to divide each number and then find the remainder individually.
Looking forward to more such techniques from you. Thanks again.
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gauri123456
@CrackVerbalGMAT- Arun Sir- Very clear. You have told a trick which will save immense time. Never knew we can solve remaiders this way. I used to divide each number and then find the remainder individually.
Looking forward to more such techniques from you. Thanks again.

:) My Pleasure. You're welcome.
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Just to add my 2 cents.
This trick is also used when addition is given instead of multiplication.

for eg, in the same question if it is given 1998+1999+2000 divided by 7. Find the remainder.
Find the individual remainder and then add them. 3+4+5 divided by 7.
So, the remainder will be 5.
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