one of my friends asked this...and I'm also curious as to what method one will use here
Q)What is the tenths digit of 39!/29!?
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method plz
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kvcpk
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What is tenths digit of 30*31*32*33*34*35*36*37*38*39aslan wrote:one of my friends asked this...and I'm also curious as to what method one will use here
Q)What is the tenths digit of 39!/29!?
32*35 = 1120
1120*30 = 33600
Hence the product is a multiple of 100.
Hence tenths digit should be 0.
Whats the OA?
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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Night reader
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39!/29!= (39*38 ... 29!)/29!= 30(9*8*7*6*5*4*3*2*1) or simply 30*9!aslan wrote:one of my friends asked this...and I'm also curious as to what method one will use here
Q)What is the tenths digit of 39!/29!?
note ... 5* ... 2 above => this gives 10 factor for the product of all numbers in factorial 9!
the ten's digit is 0
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kvcpk
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Hey NIght reader,Night reader wrote:39!/29!= (39*38 ... 29!)/29!= 30(9*8*7*6*5*4*3*2*1) or simply 30*9!aslan wrote:one of my friends asked this...and I'm also curious as to what method one will use here
Q)What is the tenths digit of 39!/29!?
note ... 5* ... 2 above => this gives 10 factor for the product of all numbers in factorial 9!
the ten's digit is 0
I think u missed something.
30*31*32...39
cannot be written as 30(9*8*7*6*5*4*3*2*1)
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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Night reader
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you are correct,kvcpk wrote:Hey NIght reader,Night reader wrote:39!/29!= (39*38 ... 29!)/29!= 30(9*8*7*6*5*4*3*2*1) or simply 30*9!aslan wrote:one of my friends asked this...and I'm also curious as to what method one will use here
Q)What is the tenths digit of 39!/29!?
note ... 5* ... 2 above => this gives 10 factor for the product of all numbers in factorial 9!
the ten's digit is 0
I think u missed something.
30*31*32...39
cannot be written as 30(9*8*7*6*5*4*3*2*1)
39*38*37*36... = (30+9)(30+8)(30+7)(30+6)...
(30+9)(30+8) foiling => 30^2 + (9+8)*30 + 9*8
(30+7)(30+6) foiling => 30^2 + (7+6)*30 + 7*6
...
(30^2+ 17*30+ 72)(30^2+ 13*30+ 42)...
30^4+ [(17*30+72)+(13*30+42)]*30 + (17*30+72)*(13*30+42)...
not the best way
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aslan
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@kvcpk:ahh..your method makes sense of taking the numbers ending in *5 and *2 their product and basically the numbers containing *0...and the ultimate product till tenths is sufficient..what if it asks ror 100th digit...I hope it doesn't do that in the 2 min question ! 
@Night _reader:I saw the foil method being used someplace..too complicated to do in the stipulated time frame and not error free
btw both got the answer right as '0'
@Night _reader:I saw the foil method being used someplace..too complicated to do in the stipulated time frame and not error free
btw both got the answer right as '0'
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kvcpk
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I dont think they would ask for hundredths digit. Even if they do, its not that difficult.aslan wrote:@kvcpk:ahh..your method makes sense of taking the numbers ending in *5 and *2 their product and basically the numbers containing *0...and the ultimate product till tenths is sufficient..what if it asks ror 100th digit...I hope it doesn't do that in the 2 min question !
@Night _reader:I saw the foil method being used someplace..too complicated to do in the stipulated time frame and not error free
btw both got the answer right as '0'
You just need to multiply the last digits of remaining numbers.
31*33*34*36*37*38*39
= 1*3*4*6*7*8*9
= 8
Hence 100ths digit is 8.
Hope this helps!!
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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Whitney Garner
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Hi aslan -aslan wrote:@kvcpk:ahh..your method makes sense of taking the numbers ending in *5 and *2 their product and basically the numbers containing *0...and the ultimate product till tenths is sufficient..what if it asks ror 100th digit...I hope it doesn't do that in the 2 min question !
@Night _reader:I saw the foil method being used someplace..too complicated to do in the stipulated time frame and not error free
btw both got the answer right as '0'
The test would be FAR more likely to ask you for the number of trailing 0s (zeros at the end of the number) in a high value factorial, or ask you for the 1000s digit (knowing that it will likely be 0 as well).
As a fun little side exercise - can you find how many trailing zeros are in 100! - you should be able to do so quite quickly
[spoiler]To find the number of trailing zeros you just need to find the total number of "10" factors in the number. That means that every 5 and every 2 would result in another 10, as would every actual 10 itself. Because there are SO many 2s in our factorial, we can just count the 5s and 10s. So let's count:
5 = (5*1) = 1 five, 15 = (5*3) = 1 five, 25 = (5*5) = 2 fives, 35 = 1 five, 45 = 1 five, 55 = 1 five, 65 = 1 five, 75 = (5*5*3) = 2 fives, 85 = 1 five, 95 = 1 five
Then the 10s
10=1 ten, 20=1 ten, 30=1 ten, 40=1 ten, 50=1 ten and 1 five, 60=1 ten, 70=1 ten, 80=1 ten, 90=1 ten, 100= 2 tens
So we just add them all up: 24 trailing 0s [/spoiler]
Whit
Whitney Garner
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Math is a lot like love - a simple idea that can easily get complicated
GMAT Instructor & Instructor Developer
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Math is a lot like love - a simple idea that can easily get complicated
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aslan
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@Whintey:thx for the info
I saw the trailing 0' somewhere else. might have been explained by Stacey.It was utilizing the same strategy of 5's and 2's...but I guess the 19! and 29!, made it a little perplexing.
btw, exactly this 'sort' of a question was featured in GRE some time back
I saw the trailing 0' somewhere else. might have been explained by Stacey.It was utilizing the same strategy of 5's and 2's...but I guess the 19! and 29!, made it a little perplexing.
btw, exactly this 'sort' of a question was featured in GRE some time back
