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Integer n is a four digit number such that it is made exactly of...... 25 May 2016, 02:32
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18% (02:30) correct 82% (02:44) wrong based on 89 sessions

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Integer n is a four digit number such that it is made exactly of 2 distinct digits. If n is divisible by 9, how many possible values does n have?

A. 59
B. 60
C. 61
D. 68
E. 69

Self-made. Suggestions welcome.

Regards,
Nalin
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KarishmaB
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Integer n is a four digit number such that it is made exactly of...... 22 Jun 2017, 02:18
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nalinnair
Integer n is a four digit number such that it is made exactly of 2 distinct digits. If n is divisible by 9, how many possible values does n have?

A. 59
B. 60
C. 61
D. 68
E. 69

Self-made. Suggestions welcome.

Regards,
Nalin

A 4 digit number made of exactly 2 distinct digits can be of 2 kinds:

- MMNN and its arrangements = 4!/2!*2! = 6
- MNNN and its arrangements = 4!/3! = 4

If the number is of the form MMNN, the sum M+N should add up to 9.
So the values can be 0,9 or 1,8... till 4,5.
But 0 is special because it cannot be in the thousands place. So the first digit will be 9 and the other 9 and two 0s will be arranged in 3!/2! = 3 ways only.
Total ways in this case = 3*1 + 6*4 = 27

If the number is of the form MNNN, M+3N should be divisible by 9. This will be divisible by 3, only if M is divisible by 3 too.
So M = 0, N = 3/6/9. Since 0 cannot be in the thousands place, there are only 3 arrangements in each case
M = 3, N = 2/5/8
M = 6, N = 1/4/7
M = 9, N = 0
If N = 0, there is only one arrangement 9000 possible.
M = 9, N = 3/6

Total ways in this case = 3*3 + 4*8 + 1 = 42

Total number of ways = 27 + 42 = 69

Answer (E)
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Integer n is a four digit number such that it is made exactly of...... 25 May 2016, 03:34
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If N is divisible by 9. The sum of N should also be divisible by 9. Therefore 2(x+y) should be divisible by 9. X and Y are the two distinct integers.

2(x+y) = 18 -----> x+y = 9.
2(x+y) = 09/27 is not possible since this would result in decimal.

x+y = 9 has 18, 27, 36, 45 combinations.

4* (4!/(2!*2!)) = 4* 6 = 24

and one more possible combination is 9,0 in which the first no should be 9. This has 3 possibilities.

so total possibilities is 24+3 = 27 :oops: :oops:

This is the answer I got. Please tell me what am I missing. :cry: :cry:
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Integer n is a four digit number such that it is made exactly of...... 25 May 2016, 03:38
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If N is divisible by 9. The sum of N should also be divisible by 9. Therefore 2(x+y) should be divisible by 9. X and Y are the two distinct integers.

2(x+y) = 18 -----> x+y = 9.
2(x+y) = 09/27 is not possible since this would result in decimal.

x+y = 9 has 18, 27, 36, 45 combinations.

4* (4!/(2!*2!)) = 4* 6 = 24

and one more possible combination is 9,0 in which the first no should be 9. This has 3 possibilities.

so total possibilities is 24+3 = 27 :oops: :oops:

This is the answer I got. Please tell me what am I missing. :cry: :cry:

You are assuming that the two distinct digits are used twice (e.g. 3366). It could be possible that one digit is used thrice and the other digit is used once (e.g. 1116)
This should help you get the answer.

Regards,
Nalin
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Integer n is a four digit number such that it is made exactly of...... 25 May 2016, 04:03
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nalinnair
premkrishna
If N is divisible by 9. The sum of N should also be divisible by 9. Therefore 2(x+y) should be divisible by 9. X and Y are the two distinct integers.

2(x+y) = 18 -----> x+y = 9.
2(x+y) = 09/27 is not possible since this would result in decimal.

x+y = 9 has 18, 27, 36, 45 combinations.

4* (4!/(2!*2!)) = 4* 6 = 24

and one more possible combination is 9,0 in which the first no should be 9. This has 3 possibilities.

so total possibilities is 24+3 = 27 :oops: :oops:

This is the answer I got. Please tell me what am I missing. :cry: :cry:

You are assuming that the two distinct digits are used twice (e.g. 3366). It could be possible that one digit is used thrice and the other digit is used once (e.g. 1116)
This should help you get the answer.

Regards,
Nalin


Then the additional combinations would be, 9000(1), 8883(4), 7776(4), 6669(4), 6660(3), 5553(4), 4446(4), 3339(4), 3330(3), 2221(4), 1116(4), 9990(3).
8*4 = 32
3*3 = 9
1*1 = 1
total = 42

Previous was 27. So total 27+42 = 69.
thank you Nalin. :) :-D :-D
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Integer n is a four digit number such that it is made exactly of...... 22 Jun 2017, 00:49
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nalinnair
Integer n is a four digit number such that it is made exactly of 2 distinct digits. If n is divisible by 9, how many possible values does n have?

A. 59
B. 60
C. 61
D. 68
E. 69

Self-made. Suggestions welcome.

Regards,
Nalin

Good question +1.

Although I got the answer, but the process is too long and the same as done by others, is something of this sort to be expected in GMAT?
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Integer n is a four digit number such that it is made exactly of...... 22 Jun 2017, 02:22
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nalinnair
Integer n is a four digit number such that it is made exactly of 2 distinct digits. If n is divisible by 9, how many possible values does n have?

A. 59
B. 60
C. 61
D. 68
E. 69

Self-made. Suggestions welcome.

Regards,
Nalin

Good question +1.

Although I got the answer, but the process is too long and the same as done by others, is something of this sort to be expected in GMAT?

No, I think it is a bit laborious. Perhaps "divisible by 5" or "is even" have a better chance.
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Integer n is a four digit number such that it is made exactly of...... 19 Apr 2021, 15:26
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nalinnair
Integer n is a four digit number such that it is made exactly of 2 distinct digits. If n is divisible by 9, how many possible values does n have?

A. 59
B. 60
C. 61
D. 68
E. 69

Self-made. Suggestions welcome.

Regards,
Nalin

A 4 digit number made of exactly 2 distinct digits can be of 2 kinds:

- MMNN and its arrangements = 4!/2!*2! = 6
- MNNN and its arrangements = 4!/3! = 4

If the number is of the form MMNN, the sum M+N should add up to 9.
So the values can be 0,9 or 1,8... till 4,5.
But 0 is special because it cannot be in the thousands place. So the first digit will be 9 and the other 9 and two 0s will be arranged in 3!/2! = 3 ways only.
Total ways in this case = 3*1 + 6*4 = 27

If the number is of the form MNNN, M+3N should be divisible by 9. This will be divisible by 3, only if M is divisible by 3 too.
So M = 0, N = 3/6/9. Since 0 cannot be in the thousands place, there are only 3 arrangements in each case
M = 3, N = 2/5/8
M = 6, N = 1/4/7
M = 9, N = 0
If N = 0, there is only one arrangement 9000 possible.
M = 9, N = 3/6

Total ways in this case = 3*3 + 4*8 + 1 = 42

Total number of ways = 27 + 42 = 69

Answer (E)

VeritasKarishma could you please explain the part in Red?
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Integer n is a four digit number such that it is made exactly of...... 19 Apr 2021, 23:52
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VeritasKarishma
nalinnair
Integer n is a four digit number such that it is made exactly of 2 distinct digits. If n is divisible by 9, how many possible values does n have?

A. 59
B. 60
C. 61
D. 68
E. 69

Self-made. Suggestions welcome.

Regards,
Nalin

A 4 digit number made of exactly 2 distinct digits can be of 2 kinds:

- MMNN and its arrangements = 4!/2!*2! = 6
- MNNN and its arrangements = 4!/3! = 4

If the number is of the form MMNN, the sum M+N should add up to 9.
So the values can be 0,9 or 1,8... till 4,5.
But 0 is special because it cannot be in the thousands place. So the first digit will be 9 and the other 9 and two 0s will be arranged in 3!/2! = 3 ways only.
Total ways in this case = 3*1 + 6*4 = 27

If the number is of the form MNNN, M+3N should be divisible by 9. This will be divisible by 3, only if M is divisible by 3 too.
So M = 0, N = 3/6/9. Since 0 cannot be in the thousands place, there are only 3 arrangements in each case
M = 3, N = 2/5/8
M = 6, N = 1/4/7
M = 9, N = 0
If N = 0, there is only one arrangement 9000 possible.
M = 9, N = 3/6

Total ways in this case = 3*3 + 4*8 + 1 = 42

Total number of ways = 27 + 42 = 69

Answer (E)

VeritasKarishma could you please explain the part in Red?

Two Cases:

1. MMNN and its arrangements (MNMN, NNMM etc)

We found 27 of these total cases.



2. MNNN and its arrangements (NNMN, NNNM etc)

M = 0, N = 3/6/9. Since 0 cannot be in the thousands place, there are only 3 arrangements in each case (NMNN, NNMN, NNNM)
So 3 arrangements for M= 0, N = 3; (3033, 3303, 3330)
3 arrangements for M= 0, N = 6;
3 arrangements for M= 0, N = 9;


M = 3, N = 2/5/8
There are 4 arrangements for each case.
M = 3, N = 2 gives 3222, 2322, 2232, 2223
Similarly 4 arrangements for M = 3, N = 5
4 arrangements for M = 3, N = 8
4 arrangements for M = 6, N = 1
4 arrangements for M = 6, N = 4
4 arrangements for M = 6, N = 7
4 arrangements for M = 9, N = 3
4 arrangements for M = 9, N = 6

If M = 9, N = 0, then only 1 arrangement is possible (9000.. rest 3 are not possible 0900, 0090 etc)

So total 42 arrangements here.

Total = 27 + 42
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