If x = 1998*1999*2000, what is the remainder when x is divided by 7?

Official Solution:

If \(x = 1998*1999*2000\), what is the remainder when \(x\) is divided by 7?

A. \(0\)
B. \(1\)
C. \(2\)
D. \(3\)
E. \(4\)


Express 1998, 1999 and 2000 as (a multiple of 7) + (something). The closest multiple of 7 to these numbers is 1995, so

\(x = (1995 + 3)*(1995 + 4)*(1995 + 5)\).

When we expand this expression, all terms there will have \(1995*\) in them (so all these terms will be divisible by 7), except the last term, which will be \(3*4*5=60\).

60 divided by 7 gives the remainder of 4.


Answer: E

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

Arun Kumar

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

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

Arun Kumar

@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.

@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.

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.