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

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Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink]
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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

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Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink]
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
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How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink]
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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]
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]
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

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Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink]
Why "B" can only take values from 10-3?
Re: How many positive 4-digit integers are divisible by 20 if the repetiti [#permalink]
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