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Math Expert Joined: 02 Aug 2009 Posts: 10679
How many 4-digit numbers can be formed from the digits 2, 3, 5, 6, 7 [#permalink] 27 Sep 2021, 18:36
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Difficulty: 15% (low)
Question Stats: 74% (00:53) correct 26% (01:20) wrong based on 65 sessionsHide Show timer Statistics
kapil1995 wrote: How many 4-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated? a) 30 b) 40 c) 60 d) 20 e) None of these For a number to be divisible by 5, it should end in 0 or 5.Thus the number will be ABC5.A, B and C can be chosen from remaining 5 numbers in 5*4*3 or 60 ways.C _________________
CrackVerbal Representative Joined: 03 Oct 2013 Affiliations: CrackVerbal Posts: 4987 Location: India
Re: How many 4-digit numbers can be formed from the digits 2, 3, 5, 6, 7 [#permalink] 27 Sep 2021, 21:19
Divisibility rule of 5: A number is divisible by 5 if the unit digit is 5 or 0Let the number be XYZWUnit place or W has only 1 choice(that is 5)Place Z has 5 choices (2,36,7,9)Place Y has 4 choices (as digits are not repeated)Place X has 3 choices(as digits are not repeated)By Fundamental Principle of Counting,total number of numbers= 1 * 5*4*3 = 60 numbers (option c) D.S
VP Joined: 16 Jun 2021 Posts: 1072
Re: How many 4-digit numbers can be formed from the digits 2, 3, 5, 6, 7 [#permalink] 27 Sep 2021, 23:03
kapil1995 wrote: How many 4-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated? a) 30 b) 40 c) 60 d) 20 e) None of these For a number to be divisible by 5 it should be having 5 or 0 at the end Therefore number of possibilities for number 5 fixed at the end and non repeating = 5*4*3*1 = 60Therefore IMO C
Re: How many 4-digit numbers can be formed from the digits 2, 3, 5, 6, 7 [#permalink] 27 Sep 2021, 23:03
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options: a) $220$ b) $249$ c) $432$ d) $216$ MyApproach: To form a 4 digit number divisible by 5 using given numbers I make cases here: Unit Digit is $0$ and other $3$ numbers can be formed in $7$ . $6$ . $5$=$210$ Unit Digit is $5$ and other $3$ numbers can be formed in $6$ . $6$ . $5$=$180$ Therefore,the required number is $390$
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120 Questions 480 Marks 120 Mins
Given: Concept used: Divisibility rule of 5: The number is divided by 5, if the unit digit of a number is either 5 or 0. Calculation: Unit digit can only be 5. There is only 1 possible way to fill unit place. Remaining places can be filled by 2, 3, 6, 7 or 9. There is 5 possible ways to fill ten's place. There is 4 possible ways to fill hundredth place as digits cannot be repeated. There is 3 possible ways to fill the first place of four digit number. Total numbers that can be formed = 1 × 5 × 4 × 3 = 60 ∴ 60 four-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9. India’s #1 Learning Platform Start Complete Exam Preparation
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