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

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Bunuel
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12 Days of Christmas GMAT Competition - Day 5: What is the remainder [#permalink] New post   18 Dec 2021, 07:00
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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



 


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Bunuel
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Re: 12 Days of Christmas GMAT Competition - Day 5: What is the remainder [#permalink] New post   19 Dec 2021, 08:00
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Bunuel wrote:
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





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Re: 12 Days of Christmas GMAT Competition - Day 5: What is the remainder [#permalink] New post   18 Dec 2021, 07:11
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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
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Re: 12 Days of Christmas GMAT Competition - Day 5: What is the remainder [#permalink] New post   18 Dec 2021, 07:14
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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
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Re: 12 Days of Christmas GMAT Competition - Day 5: What is the remainder [#permalink] New post   18 Dec 2021, 07:32
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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.
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Re: 12 Days of Christmas GMAT Competition - Day 5: What is the remainder [#permalink] New post   19 Dec 2021, 05:02
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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
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12 Days of Christmas GMAT Competition - Day 5: What is the remainder [#permalink] New post   21 Dec 2021, 03:47
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RanonBanerjee wrote:
IMO Answer D.

Last digit of each number will be remainder if we divide with 20, I.e., 222/20 remainder 2, 321/20 remainder 1....

The multiple of remainder would be 2*1*2*3*2*6*7 =1008 when divided by 20, remainder is 8.

Posted from my mobile device


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
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Re: 12 Days of Christmas GMAT Competition - Day 5: What is the remainder [#permalink] New post   26 Jan 2023, 00:21
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?
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12 Days of Christmas GMAT Competition - Day 5: What is the remainder [#permalink] New post   26 Jan 2023, 00:28
bullishdutta wrote:
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?


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.
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Re: 12 Days of Christmas GMAT Competition - Day 5: What is the remainder [#permalink] New post   10 Feb 2024, 18:57
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Re: 12 Days of Christmas GMAT Competition - Day 5: What is the remainder [#permalink]
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