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How many positive 4-digit integers are divisible by 20 if the repetiti

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Bunuel
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How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink] New post   07 Jun 2019, 02:43
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How many positive 4-digit integers are divisible by 20 if the repetition of digits is not allowed?

(A) 168
(B) 196
(C) 224
(D) 288
(E) 360
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C
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nick1816
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Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink] New post   07 Jun 2019, 03:13
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4-digit integers that are divisible by 20 can be written as 'ABCD'
'D' digit must be 0
'C' can take 4 values.(2,4,6 and 8)
'A' can take (9-1)=8 values
'B' can take (10-3)= 7 values

Total number of 4-digit integers that are divisible by 20, if repetition is not allowed= 8*7*4*1= 224


Bunuel wrote:
How many positive 4-digit integers are divisible by 20 if the repetition of digits is not allowed?

(A) 168
(B) 196
(C) 224
(D) 288
(E) 360
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Lampard42
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Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink] New post   07 Jun 2019, 03:14
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The last 2 digits must be among - 20,40,60 or 80 (00 not possible as 0 is repeated). So there are 8 options for 1st digit, 7 for 2nd digit multiplied by 4 possibility for last 2 digits. 8*7*4 = 224 IMO Option C
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Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink] New post   07 Jun 2019, 03:21
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For a number to be divisible by 20, the units digit must be 0 and the tens digit must be an even number except 0 to avoid repetition.

This way, the units digit has only 1 possibility and tens digit has 4 possibilities.

Now the thousands digit has only 8 options (1-9 except a even number that was picked as the tens digit) and the hunderds digit has 7 options

Therefore positive 4-digit integers that are divisible by 20 without repetition of digits = 8*7*4*1 = 224

Answer is (C)

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Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink] New post   07 Jun 2019, 03:54
Bunuel wrote:
How many positive 4-digit integers are divisible by 20 if the repetition of digits is not allowed?

(A) 168
(B) 196
(C) 224
(D) 288
(E) 360


last digit = 0; 1
and tens place ; 2,4,6,8; 4
hunderes ; 7 placs
thousands; 8 placses
8*7*4*1 ; 224
IMO C
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How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink] New post   07 Jul 2019, 10:17
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Can someone please explain the logic behind A being able to take only 8 values and B only 7 (when the 4-digit integer is written as ABCD). I know it has to do with the fact that repetition is not allowed but how can we logically deduce this?
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Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink] New post   07 Jul 2019, 21:36
RuZu wrote:
Can someone please explain the logic behind A being able to take only 8 values and B only 7 (when the 4-digit integer is written as ABCD). I know it has to do with the fact that repetition is not allowed but how can we logically deduce this?


RuZu Only 10 integers are there from 0,1,2,3...9. So C and D as repetition is not allowed will take 2 out. So max available for A will be 8 and B will be 7

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Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink] New post   16 Oct 2022, 19:32
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Bunuel wrote:
How many positive 4-digit integers are divisible by 20 if the repetition of digits is not allowed?

(A) 168
(B) 196
(C) 224
(D) 288
(E) 360



For any number to be divisible by 20, the last two digits should be 00, 20, 40, 60 and 80.
As repetition is not allowed we can have only 4 possibilities: 20, 40, 60 and 80.

Now, for each case, say 20, two digits gone and the A and B in ABCD can be fixed from remaining 8 digits. => 8*7

Total cases for all 4 possibilities = 8*7*4 = 224

C
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Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink] New post   23 Oct 2022, 11:28
Why "B" can only take values from 10-3?
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Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink]
23 Oct 2022, 11:28
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