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| Quantitative Questions and Answers Discuss Quantitative and other Math related questions. Post your math doubts and get it solved by the smartest brains this side of the universe ! | | | |
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29-04-2003, 08:22 AM
Quote: |
Originally Posted by icarus one more sitter on Numbers
Let S be the set of prime numbers greater than or equal to two and less than 100.
All such existing terms are then multiplied, then how many consecutive zeros would the final digit end with.
(1) 1 (2) 4
(3) 5 (4) 10
Lemme know if posting questions seperately will help the cause. Otherwise i wud compile them together and post them
cheers
Shyam | answer is 1
product of 2 and 5 can make a 0
no other prime number have 2 and 5 as factor other that 2 and 5 themselves 
PS:- posted the solution as no specification regarding posting f solutions by questioner was mentioned 
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Join Date: Mar 2003 Location: Jalandhar - Jamshedpur Age: 26 | Re: Que on Numbers.... -
29-04-2003, 08:41 AM
Quote: |
Originally Posted by icarus An interesting Question on Numbers funda..
Let D be a decimal of the form, D = 0.0a1a2a1a2a1a2……….., where digits a1 & a2 lie between 0 and 9. Then which of the following numbers necessarily produces an integer, when multiplied by D?
(1) 18 (2) 108
(3) 1980 (4) 208
Will come up with answers if there is no answer from Junta....
BFN...
shyam | 1000 D - 10 D = a1a2
D = a1a2/990
thus multiply with 1980
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02-05-2003, 06:37 PM
Quote:
icarus wrote:
An interesting Question on Numbers funda..
Let D be a decimal of the form, D = 0.0a1a2a1a2a1a2……….., where digits a1 & a2 lie between 0 and 9. Then which of the following numbers necessarily produces an integer, when multiplied by D?
(1) 18 (2) 108
(3) 1980 (4) 208
Will come up with answers if there is no answer from Junta....
BFN...
shyam
1000 D - 10 D = a1a2
D = a1a2/990
| can u pls explain 1000D - 10 D =a1a2 funda?
i did it this way:
any number divided by 11 gives reminder of the form 0.a1a2a1a2..
so if div by 110 we get 0.0a1a2a1a2...and the only choice thts a multiple of 110 is 1980
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02-05-2003, 06:47 PM
Quote: |
Originally Posted by Dan can u pls explain 1000D - 10 D =a1a2 funda? | suppose D=0.0585858585858....
now 10D=0.585858585858...
and 1000D=58.5858585858...
now 990D=58
D=58/990
hence when multiplied with 1980,it always produces an integer=116
hence works for any value of a1 and a2 i think
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Join Date: Mar 2003 Location: India | Re: Que on Numbers.... -
03-05-2003, 12:03 PM
Quote: |
Originally Posted by icarus
Let D be a decimal of the form, D = 0.0a1a2a1a2a1a2……….., where digits a1 & a2 lie between 0 and 9. Then which of the following numbers necessarily produces an integer, when multiplied by D?
(1) 18 (2) 108
(3) 1980 (4) 208
| Shyam,
First, you did not say if you want answers by PM or e-mail or on the board. I'm posting the answers here for its been long since you posted the Q.
D = 0.0a1a2a1a2a1a2...
=> 1000D = a1a2.a1a2a1a2a1a2.. - [1]
=> 10D = 0.a1a2a1a2a1a2... - [2]
[2] - [1] => 990D = a1a2
=> D has 990 in the denominator and 1980 is a multiple of 990. So, that shuld be the answer. | | | | | The Following 4 Users Say Thank You to D e e p t i For This Useful Post: | | | | | |
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Join Date: Dec 2002 Location: Chennai | Re: Number System -
04-07-2003, 07:31 PM
To find the number of different divisors of a composite number
Direct Method
Find the prime factors of the number and increase the index of each factor by 1...the continued product of the increased indices will give the result including Unity and the Number itself...
ex. for 50 and 37800
50 = 2^1 * 5^2
hence no. of divisors = (1+1) * (2+1) = 6
37800 = 2^3 * 3^3 * 5^2 * 7^1
hence no. of divisors = (3+1) * (3+1) * (2+1) * (1+1) = 96
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04-07-2003, 07:41 PM
1..the product of 2 numbers is 7168...and the HCF is 16..find the numbers...
2..a number on being divided by 5 & 7 succesively leaves the remainders 2 & 4 respectively..find the remainder when the same number is divided by 35...
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04-07-2003, 09:02 PM
answer 2) number must be in the form of
5(7x+4) + 2
= 35x +22
thus remainder = 22
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04-07-2003, 09:08 PM
answer 1)
numbers are (16,44  or (64,112)
approach is simple
devide 7168 with 16*16
we get quotient as 28
which can be represented as 1*28 or 4*7
(2*14 cannot be considered as it will give 32 hcf)
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04-07-2003, 09:09 PM
Quote: |
Originally Posted by smart_brain answer 2) number must be in the form of
5(7x+4) + 2
= 35x +22
thus remainder = 22 | answer is right...but here's a short-cut
here d1 = the first divisor = 5
r1 = the first remainder = 2
r2 = the second remainder = 4
hence the reqd remainder = (d1 * r2) + r1
hence (5*4)+2 = 22
HTH
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