## method plz

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### method plz

by aslan » Sat Dec 11, 2010 8:26 pm
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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by kvcpk » Sat Dec 11, 2010 9:42 pm
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!?
What is tenths digit of 30*31*32*33*34*35*36*37*38*39
32*35 = 1120
1120*30 = 33600
Hence the product is a multiple of 100.

Hence tenths digit should be 0.

Whats the OA?
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by Night reader » Sat Dec 11, 2010 9:43 pm
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!?
39!/29!= (39*38 ... 29!)/29!= 30(9*8*7*6*5*4*3*2*1) or simply 30*9!

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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by kvcpk » Sat Dec 11, 2010 10:14 pm
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!?
39!/29!= (39*38 ... 29!)/29!= 30(9*8*7*6*5*4*3*2*1) or simply 30*9!

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)

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by Night reader » Sun Dec 12, 2010 12:16 am
kvcpk wrote:
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!?
39!/29!= (39*38 ... 29!)/29!= 30(9*8*7*6*5*4*3*2*1) or simply 30*9!

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)
you are correct,

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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by aslan » Sun Dec 12, 2010 3:40 am
@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'

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by kvcpk » Sun Dec 12, 2010 5:30 am
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'
I dont think they would ask for hundredths digit. Even if they do, its not that difficult.
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."
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by aslan » Sun Dec 12, 2010 12:23 pm
@kvcpk:yep thanks....it hel[ps

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by Whitney Garner » Sun Dec 19, 2010 9:26 pm
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'
Hi aslan -

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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Manhattan Prep

Contributor to Beat The GMAT!

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by aslan » Sun Dec 19, 2010 11:32 pm
@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

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