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12 May 2019, 05:29
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If a number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to 43, what is the probability that this number is divisible by 11?
(A) 2/5
(B) 1/5
(C) 1/6
(D) 1/11
(E) 1/15
(A) 2/5
(B) 1/5
(C) 1/6
(D) 1/11
(E) 1/15
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12 May 2019, 06:16
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When all the digits of 5-digit integer are 9, then sum is equal to 45. We want 43, hence we can either subtract 2 from any digit or we can split it into (1,1) and subtract from any 2 digits.
Case 1. When 4 digits are 9 and 1 digit is 7
Number of integer possible- 5!/4!*1! =5
Case 2. When 3 digits are 9 and 2 digits are 8
Number of integers possible= 5!/3!*2!=10
Total five-digit numbers in which the sum of the digits is equal to 43= 10+5=15
To be divisible by 11, The difference of sum of digits at odd place and sum of digits of even places is divisible by 11
(sum of digits at odd place)- (sum of digits at even place)=11
It can't be 0 because you can't divide 43 in two equal integral parts. It can neither be 22 nor 33 in our scenario.
We know, (sum of digits at odd place)+ (sum of digits at even place)=43
Hence (sum of digits at odd place)=27 and (sum of digits at even place=16
Digits at odd places must be (9,9,9) and digits at odd places are (8,8), arrangement possible is 1 or (9,7), arrangements possible are 2
set of all five-digit numbers in which the sum of the digits is equal to 43 and also divisible by 11= 1+2=3
Probability=3/15=1/5
Case 1. When 4 digits are 9 and 1 digit is 7
Number of integer possible- 5!/4!*1! =5
Case 2. When 3 digits are 9 and 2 digits are 8
Number of integers possible= 5!/3!*2!=10
Total five-digit numbers in which the sum of the digits is equal to 43= 10+5=15
To be divisible by 11, The difference of sum of digits at odd place and sum of digits of even places is divisible by 11
(sum of digits at odd place)- (sum of digits at even place)=11
It can't be 0 because you can't divide 43 in two equal integral parts. It can neither be 22 nor 33 in our scenario.
We know, (sum of digits at odd place)+ (sum of digits at even place)=43
Hence (sum of digits at odd place)=27 and (sum of digits at even place=16
Digits at odd places must be (9,9,9) and digits at odd places are (8,8), arrangement possible is 1 or (9,7), arrangements possible are 2
set of all five-digit numbers in which the sum of the digits is equal to 43 and also divisible by 11= 1+2=3
Probability=3/15=1/5
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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