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12 Days of Christmas GMAT Competition - Day 1: What are the last two 14 Dec 2021, 07:00
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12 Days of Christmas GMAT Competition with Lots of Fun

What are the last two digits in the integer value of \(2987 *2379 * 3599 * 3025 * 4768 * 9999 * 56783\) ?

A. 50
B. 80
C. 45
D. 00
E. 90


 


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D
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jarnugirdhar
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12 Days of Christmas GMAT Competition - Day 1: What are the last two 14 Dec 2021, 07:10
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If we observe the numbers carefully:

2987∗2379∗3599∗3025∗4768∗9999∗56783

We have 3025 and 4768 in the list:

3025 has 25 as factor and 4768 has 4 as factor which implies we do have a "100" in this whole multiplication and hence, however the large multiplication result may be the "last two" digits will always be 00
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12 Days of Christmas GMAT Competition - Day 1: What are the last two 15 Dec 2021, 08:00
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Bunuel
12 Days of Christmas GMAT Competition with Lots of Fun

What are the last two digits in the integer value of \(2987 *2379 * 3599 * 3025 * 4768 * 9999 * 56783\) ?

A. 50
B. 80
C. 45
D. 00
E. 90


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12 Days of Christmas GMAT Competition - Day 1: What are the last two 14 Dec 2021, 07:05
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There are 2 5's and 2 2's there fore last 2 digits will be 00

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12 Days of Christmas GMAT Competition - Day 1: What are the last two 14 Dec 2021, 07:15
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2987 * 2379 * 3599 * 3025 * 4768 * 9999 * 56783
2987 * 2379 * 3599 * 5*5*121 * 2*2*1192 * 9999 * 56783

100( 2987 * 2379 * 121 * 1192 * 9999 * 56783)

Last two digits are 00

Ans : Option D
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12 Days of Christmas GMAT Competition - Day 1: What are the last two 14 Dec 2021, 07:22
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Bunuel
12 Days of Christmas GMAT Competition with Lots of Fun

What are the last two digits in the integer value of \(2987 *2379 * 3599 * 3025 * 4768 * 9999 * 56783\) ?

A. 50
B. 80
C. 45
D. 00
E. 90




 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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This one was easy one:
In series look for 5's and even multiples.
3025*4768
25*8 will give us 200 last two digits '00'.
No need to multiply other digits unit digit. :D

Hence D.
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12 Days of Christmas GMAT Competition - Day 1: What are the last two 14 Dec 2021, 07:30
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Bunuel

What are the last two digits in the integer value of \(2987 *2379 * 3599 * 3025 * 4768 * 9999 * 56783\) ?

A. 50
B. 80
C. 45
D. 00
E. 90

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multiplication of 3025 & 4768 for last 2 digits give 25*8=200..

multiplying XX00 with any number gives last 2 digits as 00.

Hence answer is option: D
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12 Days of Christmas GMAT Competition - Day 1: What are the last two 14 Dec 2021, 09:03
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~ nearest number
2987∗~3000
2379∗~2400
3599∗~3600
3025∗~3000
4768∗~4800
9999∗~10000
56783~56790
Since all approx nearest number end's with 00, thus end 2 digit with "00"
Option D
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12 Days of Christmas GMAT Competition - Day 1: What are the last two 14 Dec 2021, 12:07
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IMO: D

Factoring 3025 gives more than 2 powers of 5 and Factoring 4768 also gives more than 2 powers of 2. Which means the number will be multiplied with 100. Hence the last two digits will be 00.
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12 Days of Christmas GMAT Competition - Day 1: What are the last two 30 Dec 2025, 03:29
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