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30 Nov 2021, 08:24
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How many three-digit positive integers have even number of even digits?
A. \(100\)
B. \(125\)
C. \(300\)
D. \(325\)
E. \(450\)
A. \(100\)
B. \(125\)
C. \(300\)
D. \(325\)
E. \(450\)
30 Nov 2021, 08:24
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Official Solution:
How many three-digit positive integers have even number of even digits?
A. \(100\)
B. \(125\)
C. \(300\)
D. \(325\)
E. \(450\)
We are looking for three-digit positive integers which have zero or two even digits. So, we are looking for three-digit positive integers which have all odd digits or two even digits and one odd digit.
The number of three-digit positive integers with all odd digits is \(5*5*5=125\) (5 odd options for each digit: 1, 3, 5, 7, or 9).
The number of three-digit positive integers with two even digits and one odd digit is a little bit trickier to get. We have to consider two cases:
If the first digit is even (EOE or EEO), then there will be \(4*5*5+4*5*5=200\) numbers (4 even options for the first digit: 2, 4, 6, or 8 because a three-digit number cannot start with 0).
If the first digit is NOT even (OEE), then there will be \(5*5*5=125\) numbers.
The total number of such numbers is therefore, \(125+200+125=450\)
Answer: E
How many three-digit positive integers have even number of even digits?
A. \(100\)
B. \(125\)
C. \(300\)
D. \(325\)
E. \(450\)
We are looking for three-digit positive integers which have zero or two even digits. So, we are looking for three-digit positive integers which have all odd digits or two even digits and one odd digit.
The number of three-digit positive integers with all odd digits is \(5*5*5=125\) (5 odd options for each digit: 1, 3, 5, 7, or 9).
The number of three-digit positive integers with two even digits and one odd digit is a little bit trickier to get. We have to consider two cases:
If the first digit is even (EOE or EEO), then there will be \(4*5*5+4*5*5=200\) numbers (4 even options for the first digit: 2, 4, 6, or 8 because a three-digit number cannot start with 0).
If the first digit is NOT even (OEE), then there will be \(5*5*5=125\) numbers.
The total number of such numbers is therefore, \(125+200+125=450\)
Answer: E
21 May 2024, 20:58
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Answer is 325? 200+125
Bunuel
