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24 Dec 2020, 06:46
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B
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Difficulty:
45%
(medium)
Question Stats:
68% (01:49) correct
32%
(01:56)
wrong
based on 261
sessions
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How many three-digit positive integers with different digits are greater than 700?
A. 180
B. 216
C. 243
D. 300
E. 310
A. 180
B. 216
C. 243
D. 300
E. 310
Consider all positive integers with three different digits. (Note that zero cannot be the first digit.) Find the number of them which are: (a) greater than 700; (b) odd; (c) divisible by 5
24 Dec 2020, 09:22
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Bunuel
The hundreds digit can be any of the three - 7, 8 or 9. So 3 ways to choose
The tens digit can be anything but the hundreds digit. So 9 ways.
The ones digit anything but hundreds and tens digit. So 8 ways.
Total ways : 3*9*8=216
B
General Discussion
02 Mar 2021, 10:16
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Bunuel
Possible three digits are:
1. 7xx
So, number of three digits integers = 1*9(except 7)*8(except 7 and whatever at tens place)
2. 8xx
So, number of three digits integers = 1*9(except 8)*8(except 8 and whatever at tens place)
3. 9xx
So, number of three digits integers = 1*9(except 9)*8(except 9 and whatever at tens place)
Total = 9*8+9*8+9*8 = 216
Answer B.
