IIT JEE , AIEEE Forum - Mathematics , Algebra

Sushmi

1.)

Find the number of divisors of 3630 which have a remainder of 1 when divided by 4

Options :

a.) 12                          b.) 6

c.) 4                            d.) none of these

2.) Sum of all divisors of 5400 whose unit digit is 0 is:

a.) 5400    b.) 10800

c.)16800                       d.) 14400

Decoder

..so in 1)..
we want all no.s of form 4m+1..form that divide 3630..

lets factorize it..
u get 3630 = 2.3.5.11.11
now we can write
3 as 4m-1..
5 as 4n+1..
11 as 4p-1..
we want odd no. so remove two..
look..4m-1 * 4p-1..gives 4m+1 form..
(4p-1) ^2 of 11 also is 4m+1 form..so it gives this form wen power is even..
so cases..
3*5*11...
3*11..
5 only....
11^2.....
5*11^2...
and so on..

2)
u actually want factors that cointain at least one 2 and 5..
5400 = 2^3 . 3^3 . 5^2..

so include options according to condition..and find the answer..

Conjurer

1) EDIT
2) 16800, Is it right?

Decoder plz tell me the answer for the first answer.I think I have a doubt in this divisor thingy concept.

Sushmi

yeah ans is 16800....

go ahead post your process dont be afraid to express your thinking.

and thank you decoder i missed 5

. Answer to first answer is 6 silly question

.

elastiboysai

see the sum of divisors of a number of
x=p₁^ap_2^bp_3^c....p_m^r
is
p_1^a+1-1 *p_2^b+1-1 *.....
---------------------
p₁-1 p₂ -1

use dis formula here
sum of divisors ending with zero,
so choose 10 first
now u have
2^2 *3^3*5
so sum of divisors
= 10* [(2^3-1)/2-1]*{3^4-1/3-1}*(5^2-1)/5-1
=16800

elastiboysai

and as to how u get the formula look here:

Consider a number N which has been factorised as
N = x₁^p x₂^q x₃^r x_4^s...
where each x_i is prime
Now consider

(1+x₁+x₁²+...+x₁^p) (1+x₂+x₂²+...+x₂^q) (1+x₃+x₃²+...+x₃^r)(1+x₄+x₄²+...+x₄^r)
Clearly, the powers of different x_i.vary from 0 to p,q,r in each term and each term will have a combination of all the x_is
this expansion contains all divisors of the number N and hence is the sum
of all the divisors

Sushmi

Thank you very much.....good work

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