12 Days of Christmas GMAT Competition - Day 5: What is the remainder

Question

12 Days of Christmas GMAT Competition with Lots of Fun

What is the remainder when  (222 * 321 * 542 * 563 * 722 * 926 * 98347)  is divided by 20?

A. 18
B. 16
C. 12
D. 8
E. 4

gmatophobia · Accepted Answer

We can use the multiplicative rule for remainders -

(222∗321∗542∗563∗722∗926∗98347) / 20

The remainder would the product of individual remainders -

Remainder ( \frac{222}{20}  ) = 2

Remainder ( \frac{231}{20}  ) = 1

Remainder ( \frac{542}{20}  ) = 2

Remainder ( \frac{563}{20}  ) = 3

Remainder ( \frac{722}{20}  ) = 2

Remainder ( \frac{926}{20}  ) = 6

Remainder ( \frac{98347}{20}  ) = 7

The net remainder is 2 * 1 * 2 * 3 * 2 * 6 * 7 = 1008 / 20

= 8

IMO - D

brains · Answer

This can be solved by a rather intuitive method. I do not know whether this method is adopted by anyone. Here is what i did:
222∗321∗542∗563∗722∗926∗98347
We know that the above product is an even number. 
And the divisor 20 can be written as 2x10
So the above product is divisible by 2 , and thus we get

111 ∗321∗542∗563∗722∗926∗98347

Now the unit digit of the following above product is 4. 
Now this product ending with unit digit 4 when divided by 10, will have a remainder 4.

But we already divided the product by 2 in above, so this remainder has to be multiplied by 2.

So Remainder shall be 2x4 =8 , which is D in the option.

Bunuel · Answer

Expert's Global Official Video Explanation

AgYIddCXyqU

sj296 · Answer

Remainder when (222∗321∗542∗563∗722∗926∗98347) is divided by 20

=> remainder 222/20 = 2
=> 321/20 = 1
=> 542/20 = 2
=> 563/20 = 3
=> 722/20 = 2
=> 926/20 = 6
=> 98347/20 = 7

Remainder when (222∗321∗542∗563∗722∗926∗98347) is divided by 20 = Remainder when (2*1*2*3*2*6*7) is divided by 20 = Remainder when 1008 is divided by 20 = 8

OA should be D

ChakAvi · Answer

What is the remainder when  (222 * 321 * 542 * 563 * 722 * 926 * 98347)  is divided by 20?

A. 18
B. 16
C. 12
D. 8
E. 4

Dividing each no. by 20 & multiplying the remainders = 2*1*2*3*2*6*7 = 1008
1000 is divisible by 20.
So, Remainder 8
IMO Answer D

ExpertsGlobal5 · Answer

Hello  RanonBanerjee ,

Although you luckily reached the correct answer, the approach needs correction.

Dealing just the "last one digit" while finding "remainder with 20" is not correct.

Example:
23 and 33: each end with 3 but one leaves remainder 3 and the other leaves 13.

The correct approach is two deal with remainders with the "last 2 digits". See, "20" is a factor of "100" (10^2) and not "10"; so, last "2" digits must be considered.

Hope this helps.

Thanks
Experts' Global team

bullishdutta · Answer

While it intuitively makes sense to look at the last two digits, could you please explain what is the framework to decide how many digits from the end to consider?

gmatophobia · Answer

bullishdutta

Let's assume that the number is of the form  abcd

We can represent abcd as 1000*a + 100*b + 10*c + d

Note, 1000*a & 100*b is divisible by 20 as 1000 and 100 are divisible by 20.

So the number  abcd  is divisible by 20, only when 10c + d is divisible by 20.

10c + d is the last two digits of the number.

Hope this makes sense.

saynchalk · Answer

Each number is a certain value just above a multiple of 20.

222 = 2 more than a multiple of 20
321 = 1 more
542 = 2 more
563 = 3 more
722 = 2 more
926 = 6 more
98347 = 7 more

These are what will leave the remainder when divided by 20 individually. Just deal with these numbers.

(2*1*2*3*2*6*7)/20 = 1008/20 = remainder 8 (8 more than a multiple of 20)

saynchalk · Answer

Just look for  HOW MUCH  the number is  MORE THAN a MULTIPLE of the DENOMINATOR . I have posted a detailed explanation as well. It should simplify the process by a lot.

example, 222 is 2 more than a multiple of 20 (20*11 + 2)

this extra value of 2 is what will leave a remainder anyway, so just need to focus on that number

Natansha · Answer

Hi  KarishmaB  could you please help explain the below method, i dont understand why the ans is not 4, why are we again multiplying the remainder by 2 when we already took care of that 2.

Krunaal · Answer

In his first step, dividing 222 by 2 was not required, rather you just multiply the unit digits of - 222∗321∗542∗563∗722∗926∗98347 and you get 8. The correct approach is https://gmatclub.com/forum/12-days-of-c ... l#p2928005

Natansha · Answer

The approach that you have provided is very lengthy. I wanted to understand this one better, Dividing by 2 was done so that in the denominator he has only 10, which by knowing the units digit will give us the remainder, i wanted to know why the 2 was multiplied back again

12 Days of Christmas GMAT Competition - Day 5: What is the remainder

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12 Days of Christmas GMAT Competition - Day 5: What is the remainder

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