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How many positive five-digit integers contain the digit grou
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Updated on: 08 Jul 2018, 11:34
Updated on: 08 Jul 2018, 11:34
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Question Stats:
43% (01:52) correct
56% (02:31) wrong based on 58 sessions
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How many times does the digit grouping “57” (in that order) appear in all of the five-digit positive integers? For instance, “57” appears once in 12,357, twice in 57,057, and does not appear in 24,675.
(A) 279
(B) 3,000
(C) 3,500
(D) 3,700
(E) 4,000
Re: How many positive five-digit integers contain the digit grou
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21 Jun 2021, 22:30
21 Jun 2021, 22:30
2
There are four different cases that you must count: 5 7 _ _ _ ; _ 5 7 _
_ ; _ _ 5 7 _; and _ _ _ 5 7. In the case of 5 7 _ _ _, all three of the empty
spaces can have any digit from 0–9, which is 10 possibilities, for a total of 10
× 10 × 10 = 1,000 possible numbers. In the case of _ 5 7 _ _, there are only 9
choices for the first digit since you cannot put a zero there if it is to be a fivedigit positive integer. For the last two digits any number from 0–9 is still
allowed, for a total of 9 × 10 × 10 = 900 possible numbers for the second
case. The third and fourth cases are similar to the second; both include 900
possible numbers. Summing up the four cases, this adds up to 1,000 + 900(3)
= 3,700 such integers. Note that this method will double count any integer
that has two instances of the grouping “57” in it. For example, 57,357 will be
counted both in the case of 57 _ _ _ and in the case of _ _ _ 57. In total, there
are 10 ways a number could be counted both in the 57 _ _ _ and the _ _ _ 57
cases; there are 10 ways a number could be counted both in the 57 _ _ _ and
the _ _ 57 _ cases; and there are 9 ways a number could be counted both in
the _ 57 _ _ and _ _ _ 57 cases. This leaves 10 + 10 + 9 = 29 integers that are
double-counted. However, it is fine that they are double-counted, because the
question asks for the number of times the grouping “57” appears. These
integers contain that grouping twice, so they should be counted twice, and the
correct answer is 3,700.
_ ; _ _ 5 7 _; and _ _ _ 5 7. In the case of 5 7 _ _ _, all three of the empty
spaces can have any digit from 0–9, which is 10 possibilities, for a total of 10
× 10 × 10 = 1,000 possible numbers. In the case of _ 5 7 _ _, there are only 9
choices for the first digit since you cannot put a zero there if it is to be a fivedigit positive integer. For the last two digits any number from 0–9 is still
allowed, for a total of 9 × 10 × 10 = 900 possible numbers for the second
case. The third and fourth cases are similar to the second; both include 900
possible numbers. Summing up the four cases, this adds up to 1,000 + 900(3)
= 3,700 such integers. Note that this method will double count any integer
that has two instances of the grouping “57” in it. For example, 57,357 will be
counted both in the case of 57 _ _ _ and in the case of _ _ _ 57. In total, there
are 10 ways a number could be counted both in the 57 _ _ _ and the _ _ _ 57
cases; there are 10 ways a number could be counted both in the 57 _ _ _ and
the _ _ 57 _ cases; and there are 9 ways a number could be counted both in
the _ 57 _ _ and _ _ _ 57 cases. This leaves 10 + 10 + 9 = 29 integers that are
double-counted. However, it is fine that they are double-counted, because the
question asks for the number of times the grouping “57” appears. These
integers contain that grouping twice, so they should be counted twice, and the
correct answer is 3,700.
General Discussion
Re: How many positive five-digit integers contain the digit grou
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18 Sep 2017, 00:59
18 Sep 2017, 00:59
4
1
Is there a solution faster than mine?
My idea is to think that these numbers can be of three kinds, given that they are 5-digit: YXX57, YX57X or Y57XX, plus the numbers 57XXX. Then, digit Y can be each integer between 1 and 9, thus it can take nine values, whereas the X digit can take any value between 0 and 9, meaning ten values. For one of the first three numbers the possibilities are 9*10*10 = 900, thus for the three number we have a total of 2700 numbers. The numbers 57XXX instead are 10*10*10 = 1000. Summing 2700 and 1000 we get 3700, thus answer D!
My idea is to think that these numbers can be of three kinds, given that they are 5-digit: YXX57, YX57X or Y57XX, plus the numbers 57XXX. Then, digit Y can be each integer between 1 and 9, thus it can take nine values, whereas the X digit can take any value between 0 and 9, meaning ten values. For one of the first three numbers the possibilities are 9*10*10 = 900, thus for the three number we have a total of 2700 numbers. The numbers 57XXX instead are 10*10*10 = 1000. Summing 2700 and 1000 we get 3700, thus answer D!
