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Re: How many five digit positive integers are divisible by 3 [#permalink]

16 Nov 2012, 01:24

Quote:

How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated?

===========
Choices
===========
A. 15
B. 96
C. 216
D. 120
E. 625
===========

Case 1: Any 5-digit number formed using the numbers in the set {1,2,3,4,5} is divisible by 3 because 1+2+3+4+5=15 which is divisible by 3.
Thus, number of 5-digit numbers = 5! = 120

Case 2: Any 5-digit number formed using the numbers in the set {0,1,2,4,5} is divisible by 3.
Thus, number of 5-digit numbers = 5! - 4! = 96
4! is subtracted to remove all the permutations where 0 comes as the first number. For example, 01254 is not a 5-digit number. It becomes a 4-digit number. Hence, with 0 as the first number, there are 4! permutations. This should be deducted from 5!

Case 3: No other combination of 5 numbers will be divisible by 3.

From Cases 1, 2 and 3, number of 5-digit numbers = 120 + 96 = 216. (C)

Note that they have asked only for positive integers. If not, we should do 216*2
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Last edited by th03 on 16 Nov 2012, 08:00, edited 1 time in total.

Manager

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Posts: 95

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Re: Two boys begin together to write out a booklet containing 53 [#permalink]

16 Nov 2012, 01:36

Quote:

Two boys begin together to write out a booklet containing 535 lines. The first boy starts with the first line, writing at the rate of 100 lines an hour; and the second starts with the last line then writes line 534 and so on, backward proceeding at the rate of 50 lines an hour. At what line will they meet?

A) 356
B) 277
C) 357
D) 267
E) 286

Speed of first boy = 100 lines/hour
Speed of second boy = 50 lines/hour

When they meet, let time elapsed be 't' hours.

=> (100*t) + (50*t) = 535
=> t=\frac{107}{30}

In 107/30 hours time, first boy would have reached line = 100 * \frac{107}{30}
= 356.6
This means that they meet on line 357. (C)
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Last edited by th03 on 16 Nov 2012, 07:59, edited 1 time in total.

Manager

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Re: How many factors of 2^5 * 3^6 * 5^2 are perfect squares? [#permalink]

16 Nov 2012, 01:39

Vips0000 wrote:

PraPon wrote:

How many factors of 2^5 * 3^6 * 5^2 are perfect squares?

Choices:
A. 20
B. 24
C. 30
D. 36
E. 21

2^5 * 3^6 * 5^2 = 2* (2^2 * 3^3 * 5^1) ^2
=> number of perfect square factors= (2+1)*(3+1)*(1+1) = 24

Ans B it is.

i pre-phrased question incorrectly as how many perfect squares are there in 2^5 * 3^6 * 5^2
that are factors of the above number?

if answer is 24 then how come it be perfect square because perfect square have odd number of total factors.
please correct me if i am wrong.
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Re: How many factors of 2^5 * 3^6 * 5^2 are perfect squares? [#permalink]

16 Nov 2012, 02:30

Aristocrat wrote:

Vips0000 wrote:

PraPon wrote:

How many factors of 2^5 * 3^6 * 5^2 are perfect squares?

Choices:
A. 20
B. 24
C. 30
D. 36
E. 21

2^5 * 3^6 * 5^2 = 2* (2^2 * 3^3 * 5^1) ^2
=> number of perfect square factors= (2+1)*(3+1)*(1+1) = 24

Ans B it is.

i pre-phrased question incorrectly as how many perfect squares are there in 2^5 * 3^6 * 5^2
that are factors of the above number?
if answer is 24 then how come it be perfect square because perfect square have odd number of total factors.
please correct me if i am wrong.

I'm not sure what you meant in highlighted portion. Please rephrase it for me to answer.

For second part-
Ans is 24 for this question: how many factors ( of a number ) are perfect squares.
You would have been correct in objecting had the question been : how many factors are for a given number, which is a perfect square.

Hope it clarifies.
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Re: How many factors of 2^5 * 3^6 * 5^2 are perfect squares? [#permalink] 16 Nov 2012, 02:30

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