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Why2settleForLess
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Why2settleForLess
570, 572, 574, 576, and 578, so total 5. Hence Option A.

Hi,
you have fallen into trap of "IF" and a choice matching your solution....
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Hi, we also can solve this question using properties of a compliment.

Total # of 3 digit even numbers:
(# of choices for hundreds digit)*(#... for tens digit)*(#... for units digit) = 9*10*5 = 450

Now leet’s look at two cases when we break the restriction:

Case #1. 5 appears at hundreds digit place:
(# of choices for hundreds digit)*(#... for tens digit)*(#... for units digit) = 1*9 (we can’t use 7)*5 = 45

Case #2. 5 appears at tens digit place:
Same logic: 8 (can’t use 0 or 7)*1*5=40

The resultant = Total # of 3 digit even integers - # of integers, when we break the restriction = 450 – 45 – 40 = 365

Answer C.
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I think the common trap here, based on the answer distribution, is that many are forgetting the question is talking about all 3 digit even numbers (the information provided about 5 and 7 is an exception within 500 and 700 levels).
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Would anyone please explain the trap. I could not see where the except 5&7 comes in?
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Would anyone please explain the trap. I could not see where the except 5&7 comes in?


Hi...

The use of " IF 5 is there, then 7 has to be next to it...
So we can have a number having 7 but not 5..
We can also have numbers without 5 and 7..
But we cannot have a number having 5 but not 7
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chetan2u
How many 3-digit even numbers are possible such that if one of the digits is 5, the next/succeeding digit to it should be 7?
A) 5
B) 305
C) 365
D) 405
E) 495

Since the number should be even, the last place can be taken by 0,2,4,6 or 8.
Since 5 has to be followed by 7, 5 can't be the tens digit.
Knowing this information we can solve this problem:

Case 1: When 5 is present at the hundreds place:
The number will be 57_. Now _ can be filled in 5 ways - 0,2,4,6 or 8. Thus total ways=5

Case 2: When 5 is not present at the hundreds place:
The number will have 8 digits at the hundreds place (except 0 & 5).
The number will have 9 digits at the tens place (except 5).
The number will have 5 digits at the units place.
So total ways=8*9*5=360

Hence, total ways=360+5=365
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chetan2u
How many 3-digit even numbers are possible such that if one of the digits is 5, the next/succeeding digit to it should be 7?
A) 5
B) 305
C) 365
D) 405
E) 495

Since the number should be even, the last place can be taken by 0,2,4,6 or 8.
Since 5 has to be followed by 7, 5 can't be the tens digit.
Knowing this information we can solve this problem:

Case 1: When 5 is present at the hundreds place:
The number will be 57_. Now _ can be filled in 5 ways - 0,2,4,6 or 8. Thus total ways=5

Case 2: When 5 is not present at the hundreds place:
The number will have 8 digits at the hundreds place (except 0 & 5).
The number will have 9 digits at the tens place (except 5).
The number will have 5 digits at the units place.
So total ways=8*9*5=360

Hence, total ways=360+5=365
is there a reason we dont count x57 numbers?

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ok i get it even numbers. Questions like this xompletely destroy my confidence sigh.

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