A certain identification code consists of 6 digits, each of which is

Question

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

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Official Answer

C

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gmatophobia · Accepted Answer

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

GMATinsight · Answer

Please find the simple Video solution of this problem https://www.youtube.com/watch?v=yzr76Qlp230 Answer: Option C GMAT-Club-Forum-9nttr42x.png

Sonia2023 · Answer

gmatophobia  - can you please check the following solution for the question? Please let me know if this makes sense.

We are given six places wherein 5th and 6th places can be filled only in 1 way. Therefore we need to find in how many ways can first 4 places be filled.

For 1st and 3rd place = Sum of the digits can be less than or equal to 9

1st + 3rd <= 9
For making it equal, one place can be introduced - X, So, 1st + 3rd + X = 9
Now in 1st and 3rd place, the minimum number has to be 1.

So 1+1+0 = 2 minimum

Therefore, number of ways in which the 1st and 3rd places can be arranged is (7+2)!/7!*2! = 36 ways

Similarly for 2nd and 4th places, 36 ways

Given we have to form the code, we will use AND. Therefore, total ways 36*36*1*1 = 1296.

Oppenheimer1945 · Answer

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

ruis · Answer

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

KarishmaB · Answer

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)

Cuttlefish · Answer

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.

Posted from my mobile device

Posted from my mobile device

Kinshook · Answer

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

einstein801 · Answer

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

Bunuel · Answer

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.

scrantonstrangler · Answer

Whay are we discounting the cases of 0, in each case? (0+9=9) right?

Amrita01 · Answer

Since the question says each number is choosen  from 1 to 9 and not 0 to 9. So clearly you cant consider 0 here.

DanTheGMATMan · Answer

Pretty tricky, but the number of possibilites for the summed digits is really just the same of the integers 1-8:

https://www.youtube.com/watch?v=_d-LN9IF_xE

hkohn · Answer

Quick formula for the sum of the first n consecutive integers:
sum = n*(n+1)/2

or in this case

1 + 2 + 3 + ... + 8 = 8*9/2 = 36

You could also do the sum manually in this case.

Apk641 · Answer

Using Combinations

For 5th position 
9C2= 9!/2@(9-7)!= 36 ways

For 6th position 
9C2= 9!/2@(9-7)!= 36 ways

36*36= 1296

Apk641 · Answer

Step 1: Understanding the Combinations

You correctly noted that the sum of the 1st and 3rd digits determines the 5th digit, and the sum of the 2nd and 4th digits determines the 6th digit. Since we are dealing with digits from 1 to 9, we are considering combinations of two digits that sum to another digit within this same range.

To solve this, you applied the combination formula, $ \binom{n}{r} = \frac{n!}{r!(n-r)!} $, which is used to calculate how many ways you can choose $ r $ objects from a set of $ n $ without considering the order.

- For each digit, you are selecting two numbers from 1 to 9 (since the sum of two digits determines the 5th and 6th digits), and there are 9 possible digits to choose from.

Step 2: Applying Combinations for the 5th Position

For the 5th digit, which is determined by the sum of the 1st and 3rd digits, you want to calculate how many valid combinations of two numbers (1st and 3rd digits) can sum to a valid 5th digit.

The combination is calculated as:
\

This means there are 36 ways to choose the 5th digit based on the 1st and 3rd digits.

Step 3: Applying Combinations for the 6th Position

Similarly, the 6th digit is determined by the sum of the 2nd and 4th digits. Again, there are 36 valid combinations of two digits from 1 to 9 that sum to a valid 6th digit.

The combination calculation is identical to the one for the 5th position:
\

Step 4: Calculating the Total Number of Identification Codes

To find the total number of possible identification codes, you multiply the number of ways to choose the 5th and 6th digits, which are both 36:
\

Thus, there are 1,296 possible identification codes that satisfy the conditions given in the problem.

Summary:

- You correctly used the combination formula $ \binom{n}{r} $ to calculate the number of ways to choose the 5th and 6th digits.
- Each of the 5th and 6th digits has 36 possible combinations based on the sum of the respective pairs of digits.
- Multiplying these two gives the total number of valid codes: $ 36 \times 36 = 1296 $.

The answer is 1,296 identification codes.

plaxo3 · Answer

This answer doesn't make sense to me. If we can only have digits 1-9, how can we count the 5th and 6th digits where the sum equals 10 (i.e. 2 and 8, 1 and 9, etc)?

That would not be possible for the 6 digit ID since we can't use zero as a digit. This is a poorly written question.

Also if the sum if the two numbers goes over 10, it doesn't specify to only use the units digit, so that would rule out all pairs that go over (i.e. 6+7=13).

mqnycsf · Answer

I used the formula for indistinguishable items instead...

P(MMMFFF) = 6!/3!3! = 20
P(MMMFFF) = (7/14)*(6/13)*(5/12)*(7/11)*(6/10)*(5/9) ... ended up with a very long equation

I know I am wrong but can someone help me understand why I can't use this method and where did I do wrong?

braddouglas10 · Answer

O great Bunuel, the tall super attractive man, do you have any similar problems? This one seems like a complicated one for the masses! GMAT-Club-Forum-7n37x60t.png

A certain identification code consists of 6 digits, each of which is

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