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Math Expert V
Joined: 02 Aug 2009
Posts: 8008
How many 3-digit even numbers are possible  [#permalink]

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12 00:00

Difficulty:   95% (hard)

Question Stats: 26% (02:16) correct 74% (01:39) wrong based on 175 sessions

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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

_________________
Intern  Joined: 07 Feb 2016
Posts: 6
Re: How many 3-digit even numbers are possible  [#permalink]

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570, 572, 574, 576, and 578, so total 5. Hence Option A.
Math Expert V
Joined: 02 Aug 2009
Posts: 8008
How many 3-digit even numbers are possible  [#permalink]

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Why2settleForLess wrote:
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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Math Expert V
Joined: 02 Aug 2009
Posts: 8008
Re: How many 3-digit even numbers are possible  [#permalink]

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4
chetan2u wrote:
How many 3-digit even numbers are possible such that if one of the digits is 5, the next digit to it should be 7?
A) 5
B) 305
C) 365
D) 405
E) 495

OA after 3 days

Hi,
this Q can be answered correctly if we do not fall in trap of 'if'..
"if one of the digits is 5, the next digit to it should be 7" means that all other solutions are valid and only if 5 is used, we have restrictions..

the answer can be found in two different sets:-

1) when 5 is not used..

a) first digit can be any of 10 digits except 0 and 5-- so 8 ways..
b) the second digit can be any of the ten except 5===so 9 ways..
c) the third digit has to one of 5 even digits as the number is EVEN- so 5 ways.. 0,2,4,6,8

total ways= 8*9*5=360 ways...

2) when 5 is used...
a) only 5 can be used here as it is to be followed by 7 and last digit has to be even..
b) only 7 can be used
c) any of 5 even number..

total ways= 1*1*5=5 ways

TOTAL = 360 +5=365 ways
C

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Senior Manager  B
Joined: 13 Oct 2016
Posts: 359
GPA: 3.98
Re: How many 3-digit even numbers are possible  [#permalink]

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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

Current Student B
Status: DONE!
Joined: 05 Sep 2016
Posts: 357
Re: How many 3-digit even numbers are possible  [#permalink]

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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).
Intern  B
Joined: 27 Jun 2016
Posts: 2
Re: How many 3-digit even numbers are possible  [#permalink]

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Would anyone please explain the trap. I could not see where the except 5&7 comes in?
Math Expert V
Joined: 02 Aug 2009
Posts: 8008
Re: How many 3-digit even numbers are possible  [#permalink]

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ss4689 wrote:
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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Retired Moderator V
Joined: 28 Mar 2017
Posts: 1195
Location: India
GMAT 1: 730 Q49 V41 GPA: 4
Re: How many 3-digit even numbers are possible  [#permalink]

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1
chetan2u wrote:
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
_________________ Re: How many 3-digit even numbers are possible   [#permalink] 31 Jan 2018, 02:24
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