Taking 22 as the ans , the no is of form 35n+22 . Taking n as 1 , i get the no as 57. Now 57 when divided by 7 does NOT leave a remainder of 4.
IMHO the ans should be 32
What say u ???

the answer sure is 22...this is how it goes..
let A is the number which when divided by 5 gives 2 as the rem...and B is the number which when divided by 7 gives 4 as the remainder...let C be the dividend now..and for simplicity lets assume C to be 1...
hence B = 7*1 + 4 = 11
and A = 5*11 + 2 = 57
now 57%35 = 22...which is the answer!!
is this ok?

Why can't i take the first divisor as 7 and proceed as u say .
In that case I have 7*2 + 4 = 18 ?????

huh? u havent read the Qn properly...
it says "Divided Successively"...so then how can u assume 7 as the 1st divisor???? is this clear???

Why can't i take the first divisor as 7 and proceed as u say .
In that case I have 7*2 + 4 = 18 ?????

And somehow I am not convinced with the ans as 22 ( as u explained )or 18 ( as above ).
Taking 22 as the ans , the no is of form 35n+22 . Taking n as 1 , i get the no as 57. Now 57 when divided by 7 does NOT leave a remainder of 4.
IMHO the ans should be 32
What say u ???

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

1..the product of 2 numbers is 7168...and the HCF is 16..find the numbers...

to solve these kind of problems, just follow these 3 steps

1. find the value of product/(HCF^2)
2. find the possible pairs of factors of value obtained in step 1
3. multiply the HCF with the pair of prime factors obtained in step 2

same method was followed by Gautam, and the answers are correct

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

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)

answer 2) number must be in the form of
5(7x+4) + 2
= 35x +22
thus remainder = 22

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...

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

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.. -
=> 10D = 0.a1a2a1a2a1a2... -

- => 990D = a1a2
=> D has 990 in the denominator and 1980 is a multiple of 990. So, that shuld be the answer.

Dan Says

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

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 :)

Dan

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

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

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

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

. If (AB) x (AB) = XYA, then X stands for :

(1) 3 (2) 1 (3) 6 (4) 9



Hey guys,

How to go abt this?

lemme try this..
a 2-digit no. havin a 3-digit no as its square..so the 2digit no is<32.. in other words A is 1,2 or 3.
but as B^2 has its last digit as A ..therefore A cant be 2 or 3.Hence A=1..and this leads us to B=9..so 19^2=361...
And X=3..

*phew*

2. If (AB) x (AB) = XYA, then X stands for :

(1) 3 (2) 1 (3) 6 (4) 9

Hey guys,

How to go abt this?

-Natasha

yeah u r correct ........silken pointed this out to me
i was assuming it for odd powers

just forgot to edit it again
thanks

dont u think it shud be +1..??

(901 + (-1))^20/17

as (-1)^20 =1.so the remainder left is one....!!

what do ya say guys...??

why is it -1?

well its just a method to solve such questions nothing to do with remembering tables

take the powers of 30 untill you get something like 17*m +/- 1
if it wud not have been 900 .....i wud have tried representing 2700 like 17*m +/- 1

hey just wanna ask 1 question regarding this...

did these numbers strike you or u know tables pretty welll that u cud see thru the solution.

let me know...

thanks

Hi

Here's a question..I guess its quite simple...but then my maths is pathetic so never mind

Find the remainder if 30^40 is divided by 17.

Luv n Luk

30^40 /17
= 900^20/17
=(901-1)^20/17
=(53*17 - 1)^20/17

hence applying binomial theorem........ we get the remainder as 1

Hi

Here's a question..I guess its quite simple...but then my maths is pathetic so never mind

Find the remainder if 30^40 is divided by 17.

Luv n Luk

Hi

I juz happen to remember that CAT2000 had some really good question on numbers. You can find the questions in CAT2000 brochure if you have one.

Also you can juz log into PTINDIA @ http://ptindia.com/.

I found the user id and passwd some where in this forum to be

pgforum and pgforum respectively


let me know if you want me to copy paste those ques here !!!!!

Cheers
shyam

Hi

Juz found some number related probs in CAT'99 Paper. Most of the questios are sitters. I dunno if this would help the Junta

1. The difference between the biggest and the smallest number formed by using 0, 1, 2, 3, 4 exactly once is :

(1) 28000 (2) 32591 (3) 27143 (4) 41976


2. If (AB) x (AB) = XYA, then X stands for :

(1) 3 (2) 1 (3) 6 (4) 9


3. A number divided by 899 gives remainder of 63. What will be the remainder when it is divided by 29 ?

(1) 5 (2) 7 (3) 9 (4) 15

4. How many numbers can be formed by using 8,7,5,3,2 such that they are divisible by 125 ? Digits to be used exactly once.

(1) 20 (2) 4 (3) 2 (4) 1


5. If n x n x n is odd, then which is true ?
(I) n x n is even
(II) n x n is odd
(III) n is odd

(1) (I) alone (2) (I) and (II) (3) (II) and (III) (4) (II) alone

6. The unit's digit in the expansion of 2 raised to power 51 is :

(1) 2 (2) 4 (3) 6 (4) 8

7. Take a 99 digit number. It is created by using first 54 natural numbers. The number thus is 1234567891011.....5354. This number is now divided by 8. The remainder is :

(1) 1 (2) 2 (3) 4 (4) 8


8. How many 5 digit numbers can be made using 1, 2, 3, 4, 5 such that the digit in the unit place is greater than the digit in the hundred's place ?

(1) 40 (2) 120 (3) 60 (4) 720

And one or two more similar, easy questions also appeared.

thanks for that info ..

simple when the common ratio is gr8r than 1 and the series terms are increasing continously...

and btw why wud that sound dumb... no one is born with all fundaes ....:-)

mite sound dumb but have completely forgotten whts a progressive GP

Cud u clarify

GP can become AP on taking log provided we hav progressive GP....

I m sorry for missing that out...

here are the steps....
For 3 no.s a,b,c in AP, we have 2b=a+c

Hence here we have 2log(2x-1)= log2+log(2x+3)

Removing Log we have
(2x-1)^2= (2)(2x+3)

or 4x^2-8x-5=0

Solving we have x = -0.5 or 2.5
If we use x=-0.5 then second number wont exist so we have x=2.5

Now see putting it, we have three no.s as 2,4,8

Taking log we hav log2, 2log2, 3log2 ie AP...

so also note that GP can become AP on taking log provided we hav progressive GP.... :-)

Hope thats clear to everyone...

arcade79 Says

2.5

Please add the steps followed also so that others can benefit

Bharathi.

2.5