How many number of five digit can be formed with the digit 0 1 2 4 6 and 8 * 600 700 800 900
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Without considering different cases. Show
This is another way to get the same answer as already answerd above. Here we start by finding the total amount of the four-digit numbers with distinct digits, then finding the amount of odd digits, filling the same criteria, and last, we subtract the odd numbers from the total, to find the even numbers filling the criteria. We have totally seven digits. We know that the digit zero cannot be placed at the position representing the thousand position. That leaves us with six digits to choose from for this position and that can be made in $P(6,1) = \frac{6!}{(6-1)!} = \frac{6!}{5!}=6$, different ways. Now, we have three positions left to fill and six digits to choose from, including the digit zero, which can be placed anywhere in the remaining positions. This choice can be done in $P(6,3) = \frac{6!}{(6-3)!} = \frac{6!}{3!}=6 \cdot5 \cdot4$, different ways. Finally, with distinct digits, there is $6 \cdot6 \cdot5 \cdot4 =720$ four-digit numbers to be constructed. We know that the amount of odd numbers plus the amount of even numbers equal the total amount of the 720 four-digit numbers. The odd numbers are 1, 3 and 5. The question to be asked is how many of the 720 are odd? The digit to fill the unit position can only be chosen from the digits 1, 3 or 5, and this choice can be made in $P(3,1)=\frac{3!}{(3-1)!}=\frac{3!}{2!}=3$, different ways. To choose the digit filling the thousand position, we have five valid digits to choose from, since the zero digit is not valid for this position. The choice can be made in $P(5,1)=\frac{5!}{(5-1)!}=\frac{5!}{4!}=5$, different ways. Now we are left with two positions to fill, the hundred and tenth position, and five digits to choose from, now including the zero digit. The choice for these two positions can be made in $P(5,2)=\frac{5!}{(5-2)!}=\frac{5!}{3!}=5 \cdot4=20$, different ways. The total number of odd four-digit numbers, with distinct digits are $5 \cdot5 \cdot4 \cdot3 =300$. Now, we can answer the question how many of these 720, four-digit numbers, are even, by the subtraction $720-300=420$. The even numbers are 420. GMAT Club Daily PrepThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.Customized we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice we will pick new questions that match your level based on your Timer History Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.Hello Guest!It appears that you are browsing the GMAT Club forum unregistered! Signing up is free, quick, and confidential. Join 700,000+ members and get the full benefits of GMAT ClubRegistration gives you:
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Senior Manager Joined: 10 Apr 2012 Posts: 252 Location: United States Concentration: Technology, Other GPA: 2.44 WE:Project Management (Telecommunications)
How many five-digit numbers can be formed from
the digits 0, [#permalink]
00:00 Question Stats:
Hide Show timer Statistics How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4? Originally posted by
guerrero25 on 25 Jan 2014, 22:54. Edited the question and added the OA. Math Expert Joined: 02 Sep 2009 Posts: 86866
Re: How many five-digit numbers can be formed from the digits 0, [#permalink]
guerrero25 wrote: How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the
number must be divisible by 4? A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on). Intern Joined: 12 Oct 2016 Posts: 1 Location: Brazil GPA: 2.9 How many five-digit numbers can be formed from the digits 0,
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guerrero25 wrote: How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the
number must be divisible by 4? My approach: Math Expert Joined: 02 Sep 2009 Posts: 86866 Re: How many
five-digit numbers can be formed from the digits 0, [#permalink]
Bunuel wrote: guerrero25 wrote: How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no
digits can repeat and the number must be divisible by 4? A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on). Similar question to practice:
how-many-five-digit-numbers-can-be-formed-using-digits-91597.html Intern Joined: 20 Oct 2013 Posts: 41 Re: How many five-digit numbers can be formed from the digits 0, [#permalink]
Bunuel wrote: guerrero25 wrote: How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no
digits can repeat and the number must be divisible by 4? A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on). Dear Bunnel, Math Expert Joined: 02 Sep 2009 Posts: 86866 Re: How many five-digit numbers can be formed from the digits 0,
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nandinigaur wrote: Bunuel wrote: guerrero25 wrote: How many five-digit numbers can be formed from the digits
0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4? A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on). Dear Bunnel, Yes, we are told that no digits can repeat. So, if the last two digits are 04, 20, or 40, we are left with 4 digits for the first, 3 digits for the second and 2 digits for the third one. The same applies to the second case. Intern Joined: 20 Oct 2013 Posts: 41 Re: How many five-digit numbers can be formed from the digits 0,
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Bunuel wrote: nandinigaur wrote: Dear Bunnel, Yes, we are told that no digits can repeat. So, if the last two digits are 04, 20, or 40, we are left with 4 digits for the first, 3 digits for the second and 2 digits for the third one. The same applies to the second case. no what i mean is that we have 0, 1, 2, 3, 4, 5 to choose from. so, if the last two digits are 04, 20, or 40... then 3 of the 6 digits to choose from are gone. we will be left with 2,3,5??? how 4? Math Expert Joined: 02 Sep 2009 Posts: 86866 Re: How many five-digit numbers can be formed from the digits 0,
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nandinigaur wrote: Bunuel wrote: nandinigaur wrote: Dear Bunnel, Yes, we are told that no digits can repeat. So, if the last two digits are 04, 20, or 40, we are left with 4 digits for the first, 3 digits for the second and 2 digits for the third one. The same applies to the second case. no what i mean is that we have 0, 1, 2, 3, 4, 5 to choose from. so, if the last two digits are 04, 20, or 40... then 3 of the 6 digits to choose from are gone. we will be left with 2,3,5??? how 4? For EACH case of 04, 20, or 40 we used 2 digits and we are left with 4. Isn't it? VP Joined: 06 Sep 2013 Posts: 1384 Concentration: Finance
Re: How many
five-digit numbers can be formed from the digits 0, [#permalink]
To be divisible by four the number needs to end in 04, 40, 20, 12, 32 or 52. GMAT Expert Joined: 16 Oct 2010 Posts: 13203 Location: Pune, India
Re: How many 5 digit number combos are divisible by 4? [#permalink]
mitzers wrote: How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the
number must be divisible by 4? You have 6 digits and you need a 5 digit number. Karishma Intern Joined: 17 Jan 2017 Posts: 43 Location: India Concentration: General Management, Entrepreneurship GPA: 4 Re: How many five-digit numbers can be formed from the digits 0, [#permalink] Bunuel wrote: guerrero25 wrote: How many five-digit
numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4? A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on). Sir what i dont understand is
how can the 3 numbers be arranged in 4 ways in case 2? (highlighted text) Math Expert Joined: 02 Sep 2009 Posts: 86866 Re: How many five-digit numbers can be formed from the digits 0, [#permalink] srishti201996 wrote: Bunuel wrote: guerrero25 wrote: How many five-digit numbers can be formed from the digits 0,
1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4? A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on). Sir what i dont understand is how can the 3
numbers be arranged in 4 ways in case 2? (highlighted text) In that case, the last two digit can take 4 different values: 12, 24, 32, or 52. The first three digits can take 18 values. Total = 4*18. Target Test Prep Representative Joined: 14 Oct 2015 Status:Founder & CEO Affiliations: Target Test Prep Posts: 16260 Location: United States (CA) Re: How many five-digit numbers can be formed from the digits 0, [#permalink]
guerrero25 wrote: How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the
number must be divisible by 4? To be divisible by 4, the last two digits of the number must be divisible by 4. Therefore, they can be 04, 12, 20, 24, 32, 40, and 52. We can split these into two groups: 1) 04, 20, 40, and 2) 12, 24, 32, 52 See why Target Test Prep is the top rated GMAT course on GMAT Club. Read Our Reviews Intern Joined: 04 May 2021 Posts: 1 Re: How many five-digit numbers can be formed from the digits 0,
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matcarvalho wrote: guerrero25 wrote: How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if
no digits can repeat and the number must be divisible by 4? My approach: By far the BEST solution I have seen, given the time limit GMAT has, other solutions are not doable. Senior Manager Joined: 19 Oct 2014 Posts: 357 Location: United Arab Emirates
Re: How many five-digit numbers can be formed from the digits 0,
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OE: Re: How many five-digit numbers can be formed from the digits 0, [#permalink] 30 Jan 2022, 07:01 Moderators: Senior Moderator - Masters Forum 3088 posts How many 5Required number of numbers =(2×3×3×3×3)=162.
How many numbers of five digits can be formed with the digits 0 2 3 4 and 5 if the digits main repeat?Therefore, there will be 60 distinct 5 - digit numbers.
How many 5` Each place out of unit, 10th, 100th and 1000th can be filled in 5 ways.
Now form multiplication rule Total numbers `= 4 xx 5 xx 5 xx 5 xx 5 = 2500`. How many numbers of 5 digits can be formed?How Many 5-Digit Numbers are there? There are 90,000 five-digit numbers including the smallest five-digit number which is 10,000 and the largest five-digit number which is 99,999.
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