gauri123456 wrote:
@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
Arun Kumar
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Crackverbal Prep Team
www.crackverbal.com