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What is the last two digits of 65*29*37*63*71*87*62

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Bunuel
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What is the last two digits of 65*29*37*63*71*87*62 [#permalink] New post  19 Sep 2019, 03:38
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Question Stats:

47% (01:33) correct 53% (01:43) wrong based on 103 sessions
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What is the last two digits of \(65*29*37*63*71*87*62\) ?

A. 10
B. 30
C. 70
D. 90
E. 99

Show Spoiler
Find the last 2 digits
65*29*37*63*71*87*62
1. 70
2. 30
3. 10
4. 90
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CareerGeek
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What is the last two digits of 65*29*37*63*71*87*62 [#permalink] New post  Updated on: 13 Oct 2019, 03:55
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Bunuel wrote:
What is the last two digits of \(65*29*37*63*71*87*62\) ?

A. 10
B. 30
C. 70
D. 90
E. 99

Show Spoiler
Find the last 2 digits
65*29*37*63*71*87*62
1. 70
2. 30
3. 10
4. 90


65∗29∗37∗63∗71∗87∗62
= (5*13)*29*37*63*71*87*(2*31)
= 10*(13*29*37*63*71*87*31)

This question is equivalent to finding the unit of the number within the bracket and LAST TWO digits are nothing but the multiplication of unit digit and 10

So, unit digit of (13*29*37*63*71*87*31)
= 3*9*7*3*1*7*1
= 9

So, last two digits are 10*9 = 90

IMO Option D

Originally posted by CareerGeek on 19 Sep 2019, 04:49.
Last edited by CareerGeek on 13 Oct 2019, 03:55, edited 1 time in total.
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nick1816
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Re: What is the last two digits of 65*29*37*63*71*87*62 [#permalink] New post  19 Sep 2019, 04:52
\(65*29*37*63*71*87*62\)
= 5*13*29*37*63*71*87*2*31
= 10*13*29*31*37*63*71*87


Last 2 digit of 10*13*29*31*37*63*71*87
= [unit digit of(13*29*31*37*63*71*87)]*10
=90

[quote="Bunuel"]What is the last two digits of \(65*29*37*63*71*87*62\) ?

A. 10
B. 30
C. 70
D. 90
E. 99
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Re: What is the last two digits of 65*29*37*63*71*87*62 [#permalink] New post  19 Sep 2019, 09:11
since we have 5& 2 so correct option will have 0 in tens place E is out

65*29*37*63*71*87*62 ; 13*5*29*37*21*3*71*87*31*2 ; 10* last unit digit for each no (3*9*7*3*7) ; 10*9 ; 90
IMO D

Bunuel wrote:
What is the last two digits of \(65*29*37*63*71*87*62\) ?

A. 10
B. 30
C. 70
D. 90
E. 99

Show Spoiler
Find the last 2 digits
65*29*37*63*71*87*62
1. 70
2. 30
3. 10
4. 90
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rishab0507
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Re: What is the last two digits of 65*29*37*63*71*87*62 [#permalink] New post  19 Sep 2019, 11:29
i have other way to come to answer
As we spot 5 and 2, we know last digit is 0 , So 65*62 = 4030, So 0 is last and take 3 from this and multiply with last digit of all other


65∗29∗37∗63∗71∗87∗62

So we have last digits of remaining numbers
3*9*7*3*1*7 = --49, So take 9 and we already have 0
Answer 90
Just make sure here you don't take 65 and 62 again
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Re: What is the last two digits of 65*29*37*63*71*87*62 [#permalink] New post  23 Aug 2023, 08:04
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Re: What is the last two digits of 65*29*37*63*71*87*62 [#permalink]
23 Aug 2023, 08:04
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