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How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2

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How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2  [#permalink]

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08 Apr 2016, 01:48
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How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2?

A. 2,000
B. 1,900
C. 1,800
D. 1,600
E. 1,500

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Re: How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2  [#permalink]

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08 Apr 2016, 02:08
Digit A can be filled in 9 ways (excluding 0)
Digit B can be filled in 10 ways
Digit C can be filled in 10 ways
Digit D can be filled in 2 ways (because of modulus A-D )

The answer will be 9*10*10*2 = 1800

Ans C
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How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2  [#permalink]

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08 Apr 2016, 03:53
raarun wrote:
Digit A can be filled in 9 ways (excluding 0)
Digit B can be filled in 10 ways
Digit C can be filled in 10 ways
Digit D can be filled in 2 ways (because of modulus A-D )

The answer will be 9*10*10*2 = 1800

Ans C

hi,
I donot think C should be the answer..
In highlighted portion you have taken following extra ways
1) A=0 and D-8
2) A and D as 1 and 9..so 3*100= 300 extra ways ans 1800-300 = 1500.

we are looking for |A-D|=2 so any of the two can be bigger..
these two can have values (1,3) ; (2,4) so on till (7,9) so total 7*2 = 14..
another arrangement could be A as 2 and D as 0... D as 2 and A as 0 is not possible as number will become 3 digits..
so the ways A and D can be placed = 14+1=15 ways..
B and C can be placed in 10*10 ways..

Total = 15*10*10=1500
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Re: How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2  [#permalink]

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21 Apr 2016, 23:58
Ans = 1600 (D)

(1,3)(2,4)...(7,9) = 7*2*10*10 = 1400
(2,0), (8,0) = 2*1*10*10 = 200

=> Ans = 1600
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Re: How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2  [#permalink]

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22 Apr 2016, 01:22
1
chetan2u wrote:

hi,
I donot think C should be the answer..
In highlighted portion you have taken following extra ways
1) A=0 and D-8
2) A and D as 1 and 9..so 3*100= 300 extra ways ans 1800-300 = 1500.

we are looking for |A-D|=2 so any of the two can be bigger..
these two can have values (1,2) ; (2,4) so on till (7,9) so total 7*2 = 15..
another arrangement could be A as 8 and D as 0... D as 8 and A as 0 is not possible as number will become 3 digits..
so the ways A and D can be placed = 14+1=15 ways..
B and C can be placed in 10*10 ways..

Total = 15*10*10=1500

Hi chetan2u,

A few observations, highlighted portion might be a typo.
It should read: (1,3) ; (2,4) so on till (7,9) so total 7*2 = 14

Also, If A = 8 and D = 0, then |A-D| = 8 and not 2
Here D = 0, not 10
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Re: How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2  [#permalink]

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23 Apr 2016, 06:50
rathan1488 wrote:
Ans = 1600 (D)

(1,3)(2,4)...(7,9) = 7*2*10*10 = 1400
(2,0), (8,0) = 2*1*10*10 = 200

=> Ans = 1600

i think 8 cannot be consider with 0 their mod diff is 8 not 2 hence 1500. Please correct if i calculated anything wrong
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Re: How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2  [#permalink]

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23 Apr 2016, 07:38
Bunuel wrote:
How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2?

A. 2,000
B. 1,900
C. 1,800
D. 1,600
E. 1,500

Good one...

ABCD is a four digit number , so A must be a non zero digit

Further |A – D| = 2

Lets draw a table
Attachment:

A - D Table.PNG [ 9.21 KiB | Viewed 2375 times ]

So we have A-D 15 possible arrangements

Now go for B & C Possible arrangements
Attachment:

B - C.PNG [ 1.52 KiB | Viewed 2370 times ]

For B = The digit can be any number from 0 - 9 ( 10 digits )

For C = The digit can be any number from 0 - 9 ( 10 digits )

So, The possible arrangements is 15 x 10 x 10 = 1500
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Re: How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2  [#permalink]

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23 Apr 2016, 09:53
OptimusPrepJanielle wrote:
chetan2u wrote:

hi,
I donot think C should be the answer..
In highlighted portion you have taken following extra ways
1) A=0 and D-8
2) A and D as 1 and 9..so 3*100= 300 extra ways ans 1800-300 = 1500.

we are looking for |A-D|=2 so any of the two can be bigger..
these two can have values (1,2) ; (2,4) so on till (7,9) so total 7*2 = 15..
another arrangement could be A as 8 and D as 0... D as 8 and A as 0 is not possible as number will become 3 digits..
so the ways A and D can be placed = 14+1=15 ways..
B and C can be placed in 10*10 ways..

Total = 15*10*10=1500

Hi chetan2u,

A few observations, highlighted portion might be a typo.
It should read: (1,3) ; (2,4) so on till (7,9) so total 7*2 = 14

Also, If A = 8 and D = 0, then |A-D| = 8 and not 2
Here D = 0, not 10

Hi there were typos corrected.. Thank you
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Re: How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2  [#permalink]

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29 Nov 2019, 08:28
Bunuel wrote:
How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2?

A. 2,000
B. 1,900
C. 1,800
D. 1,600
E. 1,500

Let's make the difference between A and D as 2.
First lets consider evens:
2 0
0 2
2 4
4 2
6 4
4 6
8 6
6 8

Odds:
1 3
3 1
3 5
5 3
7 5
5 7
9 7
7 9

All Except 0 2 don't work because that won't give us a 4 digit number.

Leaving 0,2 we have 15 choice. That's a straight E.

But to actually calculate, A and D at extremes can take 15 values, The middle two digits can take all 10 digits possible So we have
15*10*10

1500
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Re: How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2   [#permalink] 29 Nov 2019, 08:28
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