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How many 3-digit even numbers are possible

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How many 3-digit even numbers are possible [#permalink]

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New post 29 Feb 2016, 03:47
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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
[Reveal] Spoiler: OA

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Re: How many 3-digit even numbers are possible [#permalink]

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New post 29 Feb 2016, 06:41
570, 572, 574, 576, and 578, so total 5. Hence Option A.
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New post 29 Feb 2016, 19:26
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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Re: How many 3-digit even numbers are possible [#permalink]

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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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Re: How many 3-digit even numbers are possible [#permalink]

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New post 15 Nov 2016, 02:15
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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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Re: How many 3-digit even numbers are possible [#permalink]

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New post 18 Nov 2016, 19:55
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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Re: How many 3-digit even numbers are possible [#permalink]

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New post 30 Jan 2018, 20:08
Would anyone please explain the trap. I could not see where the except 5&7 comes in?
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Re: How many 3-digit even numbers are possible [#permalink]

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New post 30 Jan 2018, 20:55
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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Re: How many 3-digit even numbers are possible [#permalink]

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New post 31 Jan 2018, 01:24
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
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Re: How many 3-digit even numbers are possible   [#permalink] 31 Jan 2018, 01:24
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