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

Question

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

nick1816 · Accepted Answer

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

firas92 · Answer

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)

Lampard42 · Answer

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

Archit3110 · Answer

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

RuZu · Answer

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?

Lampard42 · Answer

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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chetan2u · Answer

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

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