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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7738
GMAT 1: 760 Q51 V42 GPA: 3.82
How many 4-digit numbers greater than 3,000 have the digits: 1, 3, 5,  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 66% (01:36) correct 34% (01:25) wrong based on 87 sessions

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[Math Revolution GMAT math practice question]

How many $$4$$-digit numbers greater than $$3,000$$ have the digits: $$1, 3, 5,$$ and $$7$$?

$$A. 6$$
$$B. 9$$
$$C. 12$$
$$D. 15$$
$$E. 18$$

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Re: How many 4-digit numbers greater than 3,000 have the digits: 1, 3, 5,  [#permalink]

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2
1
MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many $$4$$-digit numbers greater than $$3,000$$ have the digits: $$1, 3, 5,$$ and $$7$$?

$$A. 6$$
$$B. 9$$
$$C. 12$$
$$D. 15$$
$$E. 18$$

MathRevolution question does not specify whether digits can be repeated or not ; seems answer is co relating to the latter...
since we have 4 digits
and >3000 so
first digit can be either 3,5,7 only '3' possibilities ;
second digit : 3 possibliites ; could be any of
third digit 2 option and 4th option 1

so 3*3*2*1 = 18 IMO E
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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7738
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: How many 4-digit numbers greater than 3,000 have the digits: 1, 3, 5,  [#permalink]

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=>

We need to count the $$4$$-digit numbers with thousands digits $$3, 5$$ and $$7$$.
The number of $$4$$-digit numbers beginning with $$3$$ is $$6$$.
The number of $$4$$-digit numbers beginning with $$5$$ is $$6$$.
The number of $$4$$-digit numbers beginning with $$7$$ is $$6$$.
Thus, the total number of such $$4$$-digit numbers is $$18 = 6 + 6 + 6.$$

Therefore, the answer is E.
Answer: E
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Manager  B
Joined: 11 Jun 2018
Posts: 96
GMAT 1: 500 Q39 V21 Re: How many 4-digit numbers greater than 3,000 have the digits: 1, 3, 5,  [#permalink]

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chetan2u the question does not specify if the digits are repeating or not

First digit can be taken in 3 ways from 3,5,7
Second digit can be selected in 4 ways from 1,3,5,7
Third digit can be selected in 4 ways from 1,3,5,7
Fourth digit can be selected in 4 ways from 1,3,5,7

If it was mentioned the digits cannot be repeating then,

First digit can be selected in 3 ways from 3,5,7
Second digit can be selected in 3 ways from remaining numbers
Third digit can be selected in 2 ways from remaining numbers
Fourth digit can be selected in 1 way

Which gives 18, but shouldn't it be mentioned in the question?
Math Expert V
Joined: 02 Aug 2009
Posts: 7756
Re: How many 4-digit numbers greater than 3,000 have the digits: 1, 3, 5,  [#permalink]

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1
Manat wrote:
chetan2u the question does not specify if the digits are repeating or not

First digit can be taken in 3 ways from 3,5,7
Second digit can be selected in 4 ways from 1,3,5,7
Third digit can be selected in 4 ways from 1,3,5,7
Fourth digit can be selected in 4 ways from 1,3,5,7

If it was mentioned the digits cannot be repeating then,

First digit can be selected in 3 ways from 3,5,7
Second digit can be selected in 3 ways from remaining numbers
Third digit can be selected in 2 ways from remaining numbers
Fourth digit can be selected in 1 way

Which gives 18, but shouldn't it be mentioned in the question?

Yes, the language could have been slightly better.
It could be ' How many 4-digit number greater than 3000 can be formed from digits 1, 3, 5 and 7 without repetition.

But, by using AND and the kind of wording, I would take it that the intention is to find the numbers containing all the digits 1, 3, 5 and 7.
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Intern  B
Joined: 23 Oct 2013
Posts: 44
Re: How many 4-digit numbers greater than 3,000 have the digits: 1, 3, 5,  [#permalink]

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but its mentioned in question whether repeatation is allowed or not

if repetation is allowed ans would be 192
Target Test Prep Representative D
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Re: How many 4-digit numbers greater than 3,000 have the digits: 1, 3, 5,  [#permalink]

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many $$4$$-digit numbers greater than $$3,000$$ have the digits: $$1, 3, 5,$$ and $$7$$?

$$A. 6$$
$$B. 9$$
$$C. 12$$
$$D. 15$$
$$E. 18$$

We see that we have 3 options for the thousands digit, 3 for the hundreds digit, 2 for the tens digit and 1 for the units digit. Thus, the total number of options is 3 x 3 x 2 x 1= 18.

Answer: E
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