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Re: A certain identification code consists of 6 digits, each of which is [#permalink]
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let the code be \(ABCDEF\)
\(a+c=2-9\) & \(b+d=2-9\)
Req ans=(No of +ve integer sol)^2 =\((1C1+2C1+3C1…+8C1)^2=1296\)
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Re: A certain identification code consists of 6 digits, each of which is [#permalink]
gmatophobia wrote:
Bunuel wrote:
A certain identification code consists of 6 digits, each of which is chosen from 1, 2, ..., 9. The same digit can appear more than once. If the sum of the 1st and 3rd digits is the 5th digit, and the sum of the 2nd and 4th digits is the 6th digit, how many identification codes are possible?

A. 625
B. 720
C. 1,296
D. 2,025
E. 6,561


A good one !

The fifth and the sixth place can be filled only in one way.

Let's take the first and the third position -

Assume we have the number \(1\) in the first digit, the third digit can be {\(1,2,3,4,5,6,7,8\)}. Note: The third digit cannot be the number \(9\), else the sum of the first and the third digit will be \(10\), and the fifth digit will be a two-digit number. This is not possible.

If we have the number \(2\) in the first digit, the third digit can be {\(1,2,3,4,5,6,7\)}. Note: The third digit cannot be the number \(8\) or \(9\), else the sum of the first and the third digit will be a two-digit number. This is not possible.

If we have the number \(3\) in the first digit, the third digit can be {\(1,2,3,4,5,6\)}. Note: The third digit cannot be the number \(7\) or \(8\) or \(9\), else the sum of the first and the third digit will be a two-digit number. This is not possible.

So we see a pattern here

First Digit → The number of ways the third digit can be filled in

If the first digit is = 1 → The number of ways the third digit can be filled in = 8
If the first digit is = 2 → The number of ways the third digit can be filled in = 7
If the first digit is = 3 → The number of ways the third digit can be filled in = 6
If the first digit is = 4 → The number of ways the third digit can be filled in = 5
If the first digit is = 5 → The number of ways the third digit can be filled in = 4
If the first digit is = 6 → The number of ways the third digit can be filled in = 3
If the first digit is = 7 → The number of ways the third digit can be filled in = 2
If the first digit is = 8 → The number of ways the third digit can be filled in = 1
If the first digit is = 9 → The number of ways the third digit can be filled in = 0 (ignore this case)

Hence, we can have \(8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36\) pairs of the first and the third digit.

The same analysis also holds true for the second and the fourth digit.

Hence, there will be 36 pairs of the second and the fourth digit as well.

Number of possible identification codes =\( 36 * 36 = 1296\)

Option C

Is there another approach to do this in a faster way? Perhaps by answer elimination? For example, we directly eliminate option B because we know the answer is not 6! And we eliminate option E because we know it is not 9^4. What about A and D? Any red flags?

My second question is, if we are talking about codes (order is important) and that the same letter appears more than once (there can be identicals), it should be a Permutation right? I do not understand why it is a Combination...

reto ScottTargetTestPrep
Bunuel KarishmaB
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Re: A certain identification code consists of 6 digits, each of which is [#permalink]
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ruis wrote:
gmatophobia wrote:
Bunuel wrote:
A certain identification code consists of 6 digits, each of which is chosen from 1, 2, ..., 9. The same digit can appear more than once. If the sum of the 1st and 3rd digits is the 5th digit, and the sum of the 2nd and 4th digits is the 6th digit, how many identification codes are possible?

A. 625
B. 720
C. 1,296
D. 2,025
E. 6,561


A good one !

The fifth and the sixth place can be filled only in one way.

Let's take the first and the third position -

Assume we have the number \(1\) in the first digit, the third digit can be {\(1,2,3,4,5,6,7,8\)}. Note: The third digit cannot be the number \(9\), else the sum of the first and the third digit will be \(10\), and the fifth digit will be a two-digit number. This is not possible.

If we have the number \(2\) in the first digit, the third digit can be {\(1,2,3,4,5,6,7\)}. Note: The third digit cannot be the number \(8\) or \(9\), else the sum of the first and the third digit will be a two-digit number. This is not possible.

If we have the number \(3\) in the first digit, the third digit can be {\(1,2,3,4,5,6\)}. Note: The third digit cannot be the number \(7\) or \(8\) or \(9\), else the sum of the first and the third digit will be a two-digit number. This is not possible.

So we see a pattern here

First Digit → The number of ways the third digit can be filled in

If the first digit is = 1 → The number of ways the third digit can be filled in = 8
If the first digit is = 2 → The number of ways the third digit can be filled in = 7
If the first digit is = 3 → The number of ways the third digit can be filled in = 6
If the first digit is = 4 → The number of ways the third digit can be filled in = 5
If the first digit is = 5 → The number of ways the third digit can be filled in = 4
If the first digit is = 6 → The number of ways the third digit can be filled in = 3
If the first digit is = 7 → The number of ways the third digit can be filled in = 2
If the first digit is = 8 → The number of ways the third digit can be filled in = 1
If the first digit is = 9 → The number of ways the third digit can be filled in = 0 (ignore this case)

Hence, we can have \(8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36\) pairs of the first and the third digit.

The same analysis also holds true for the second and the fourth digit.

Hence, there will be 36 pairs of the second and the fourth digit as well.

Number of possible identification codes =\( 36 * 36 = 1296\)

Option C

Is there another approach to do this in a faster way? Perhaps by answer elimination? For example, we directly eliminate option B because we know the answer is not 6! And we eliminate option E because we know it is not 9^4. What about A and D? Any red flags?

My second question is, if we are talking about codes (order is important) and that the same letter appears more than once (there can be identicals), it should be a Permutation right? I do not understand why it is a Combination...

reto ScottTargetTestPrep
Bunuel KarishmaB
Kinshook Archit3110 abhimahna chetan2u



I wouldn't focus on numerical answer options and to be honest, I wouldn't be able to identify that 9^4 = 6561. It is not a number I come across in GMAT much.
As for solving the question, I see that sum of 1st and 3rd digits is the 5th digit which means that sum of 1st and 3rd digits much be 9 or less.
If the 1st digit is 1, the 3rd digit could be anything from 1 to 8.
If the 1st digit is 2, the 3rd digit could be anything from 1 to 7.
and so on... We see the pattern here.
The first digit could be anything from 1 to 8 and the second digit would take 8 values or 7 values or 6 values ... 1 value.
Hence, we get 8 + 7 + ... + 1 = 8*9/2 = 36 possible values for 1st and 3rd digits. The 5th digit is defined once we pick values for 1st and 3rd digits.

The same will be applicable to 2nd, 4th and 6th digits too.

So number of ways = 36 * 36 which ends with a 6 and hence answer is (C)
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Re: A certain identification code consists of 6 digits, each of which is [#permalink]
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I interpreted 36*36 using sets theory rather than combinations. We have 36 pairs in the first set (1st and 3rd numbers) and 36 pairs in the second set (2nd and 4th numbers). Total ways taking 1 pair per set is therefore 36*36 according to the multiplication rule for sets.

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Re: A certain identification code consists of 6 digits, each of which is [#permalink]
Given: A certain identification code consists of 6 digits, each of which is chosen from 1, 2, ..., 9. The same digit can appear more than once.
Asked:  If the sum of the 1st and 3rd digits is the 5th digit, and the sum of the 2nd and 4th digits is the 6th digit, how many identification codes are possible?

Let the identification code be abcdef ; where a,b,c,d,e,f are digits in their respective places.
a + c = e
b + d = f
Since 1<={a,b,c,d,e,f} <=9
2<= a + c = e <=9
(a,c) = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(2,7),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(5,1),(5,2),(5,3),(5,4),(6,1),(6,2),(6,3),(7,1),(7,2),(8,1)}

Number of possible cases for (a,c) = 8+7+6+5+4+3+2+1 = 36 and after that e=a+c has only 1 value
Similarily
Number of possible cases for (b,d) = 8+7+6+5+4+3+2+1 = 36 and after that f=b+d has only 1 value

Total numer of such cases = 36*36 = 1296

IMO C
­
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Re: A certain identification code consists of 6 digits, each of which is [#permalink]
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Hello, I got the 36 options but actually did it as 2*36*2*36. The reason for the *2 is because positions 1,3 and 2,4 can be switched. Could anyone explain the flaw in my logic please? Thank you
gmatophobia wrote:
Bunuel wrote:
A certain identification code consists of 6 digits, each of which is chosen from 1, 2, ..., 9. The same digit can appear more than once. If the sum of the 1st and 3rd digits is the 5th digit, and the sum of the 2nd and 4th digits is the 6th digit, how many identification codes are possible?

A. 625
B. 720
C. 1,296
D. 2,025
E. 6,561

A good one !

The fifth and the sixth place can be filled only in one way.

Let's take the first and the third position -

Assume we have the number \(1\) in the first digit, the third digit can be {\(1,2,3,4,5,6,7,8\)}. Note: The third digit cannot be the number \(9\), else the sum of the first and the third digit will be \(10\), and the fifth digit will be a two-digit number. This is not possible.

If we have the number \(2\) in the first digit, the third digit can be {\(1,2,3,4,5,6,7\)}. Note: The third digit cannot be the number \(8\) or \(9\), else the sum of the first and the third digit will be a two-digit number. This is not possible.

If we have the number \(3\) in the first digit, the third digit can be {\(1,2,3,4,5,6\)}. Note: The third digit cannot be the number \(7\) or \(8\) or \(9\), else the sum of the first and the third digit will be a two-digit number. This is not possible.

So we see a pattern here

First Digit → The number of ways the third digit can be filled in

If the first digit is = 1 → The number of ways the third digit can be filled in = 8
If the first digit is = 2 → The number of ways the third digit can be filled in = 7
If the first digit is = 3 → The number of ways the third digit can be filled in = 6
If the first digit is = 4 → The number of ways the third digit can be filled in = 5
If the first digit is = 5 → The number of ways the third digit can be filled in = 4
If the first digit is = 6 → The number of ways the third digit can be filled in = 3
If the first digit is = 7 → The number of ways the third digit can be filled in = 2
If the first digit is = 8 → The number of ways the third digit can be filled in = 1
If the first digit is = 9 → The number of ways the third digit can be filled in = 0 (ignore this case)

Hence, we can have \(8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36\) pairs of the first and the third digit.

The same analysis also holds true for the second and the fourth digit.

Hence, there will be 36 pairs of the second and the fourth digit as well.

Number of possible identification codes =\( 36 * 36 = 1296\)

Option C

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Re: A certain identification code consists of 6 digits, each of which is [#permalink]
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unicornilove wrote:
Hello, I got the 36 options but actually did it as 2*36*2*36. The reason for the *2 is because positions 1,3 and 2,4 can be switched. Could anyone explain the flaw in my logic please? Thank you

­

­
36 combinations include both (1, 2) and (2, 1) pairs.

If the 1st digit is 1, the 3rd digit could be anything from 1 to 8, giving pairs such as (1, 1), (1, 2), (1, 3), ..., (1, 8)
If the 1st digit is 2, the 3rd digit could be anything from 1 to 7, giving pairs such as (2, 1), (2, 2), (2, 3), ..., (2, 7)
...
If the 1st digit is 8, the 3rd digit could only be 1, giving (8, 1) pair.­
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