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28 Dec 2021, 09:15
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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
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(A) 2/5
(B) 1/5
(C) 1/6
(D) 1/11
(E) 1/15
Are You Up For the Challenge: 700 Level Questions
29 Dec 2021, 00:51
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Bunuel
The maximum SUM of digits of a 5-digit number is 9*5 or 45 when the number is 99999.
So the sum can be 43 in only two cases:-
1) Number is made from digits 9, 9, 9, 9 and 7.
7 could be at any of the 5 places and remaining 9 can be placed at other 4 places. => Total ways = \(\frac{5!}{4!}=5\)
Say, 7 is at odd places - 79999 or 99799 or 99997. In each of the cases, difference in sum of digits at even and odd places is 9+9+7-(9+9) = 7, which is NOT divisible by 11.
Now, if 7 is at even places - 97999 or 99979. In each of the cases, difference in sum of digits at even and odd places is 9+9+9-(9+7) = 11, which is divisible by 11.
So, 97999 and 99979 are two such cases.
2) Number is made from digits 9, 9, 9, 8 and 8.
=> Total ways = \(\frac{5!}{3!2!}=10\)
Say, 8 is at odd places - 89899 or 99898 or 89998. In each of the cases, difference in sum of digits at even and odd places is 9+8+8-(9+9) = 7, which is NOT divisible by 11.
Now, if 8 is at even places - 98989 . In this case, difference in sum of digits at even and odd places is 9+9+9-(8+8) = 11, which is divisible by 11.
Also, if 8 is at both even and odd place for example - 98899 . In this case, difference in sum of digits at even and odd places is 9+8+9-(9+8) = 9, which is NOT divisible by 11.
So, 98989 is another such case.
Total 3 cases out of possible 10+5 or 15 cases => P = \(\frac{3}{15}=\frac{1}{5}\)
B
04 Sep 2024, 11:44
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