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# The first four digits of the six-digit initial password for a shopper'

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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]
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The sum of 1st and 3rd digit is 3, third digit can't be more than 1
All digits needs to be positive integers so the only possible value seems 2,1 so 1st Digit would be 2 and 3rd would be 1
And 4th Digit is easily figurable

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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]
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itoyj wrote:

The sum of 1st and 3rd digit is 3, third digit can't be more than 1
All digits needs to be positive integers so the only possible value seems 2,1 so 1st Digit would be 2 and 3rd would be 1
And 4th Digit is easily figurable

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the red part is not correct. C can be 0 or 1
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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]

The sum of 1st and 3rd digit is 3, third digit can't be more than 1
[color=#ff0000]All digits needs to be positive integers[/color] so the only possible value seems 2,1 so 1st Digit would be 2 and 3rd would be 1
And 4th Digit is easily figurable

[size=80][b][i]Posted from my mobile device[/i][/b][/size][/quote]

the red part is not correct. C can be 0 or 1[/quote]

Isn't it obvious, that dates needs to be positive, I thought that it is not even assumption, it is universal truth
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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]
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Dear itoyj
dates needs to be positive, but the tens digit of it can be zero.
if CD is the two-digit representing the month,
then January is 01 (where C = 0 and D = 1)
and December is 12 (where C = 1 and D = 2)
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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]
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MahmoudFawzy wrote:
let the -digit be ABCD16

if (A+C)*7 produce a number with a unit digit of 1, then A+C = 3,
and the possible pairs of (A,C) are (3,0) and (2,1) because C can only be 0 or 1 (CD is the month which range between 01 and 12)

if (B+D)*3 produce a number with a unit digit of 6, then B+D can be equal to 2 or 8 , which produce too many possibilities.

from statement (1), the number is 21C116
as we deduced, if A = 2, then C = 1 , and the month is 11 (November) --> sufficient

from statement (2), the number is A1CD16 , and we already know that A+C = 3 --> insufficient
if B = 1, then D can be 1 or 7, and C can be 1 or 0 respectively

B+D = 8 and 8*3=24. That doesn't have a units digit of 6. I disagree with the red highlighted part
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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]
gmatt1476 wrote:
The first four digits of the six-digit initial password for a shopper's card at a certain grocery store is the customer's birthday in day-month digit form. For example, 15 August corresponds to 1508 and 5 March corresponds to 0503. The 5th digit of the initial password is the units digit of seven times the sum of the first and third digits, and the 6th digit of the initial password is the units digit of three times the sum of the second and fourth digits. What month, and what day of that month, was a customer born whose initial password ends in 16 ?

(1) The customer's initial password begins with 21, and its fourth digit is 1.
(2) The sum of the first and third digits of the customer's initial password is 3, and its second digit is 1.

DS21891.01

So basically the question says,
There's a six digit number. What are the first, second, third and fourth digits?

Let First digit = f
Second digit = s
Third digit..... = t
Fourth digit .. = x
Fifth digit...... = 1
Sixth digit ..... = 6

And given in the question stem also is that;
1 = 7(f + t) Note: 1 is the unit digit

6 = 3(s + x) Note: 6 is the unit digit

You're asked to find the values of each of f, s, t, x!

(1) Tells us that f, s, x are 2,1,1 respectively, leaving us with only t missing.
Using 1 = 7(f + t) ---->>>> 1 = 7(1 + t); so only one unknown in one equation tells you the equation is solvable. No need to solve! No time! (1) is SUFFICIENT.

(2) Tells us that f + t = 3 and s = 1. We are finding f, t, x!
So using... 6 = 3(s + x) and 1 = 7(f + t)

6 = 3(1+x) solvable! now let's see if we can find f, t
1 = 7(f + t)
3 = f + t
Looks like two equations with 2 variables but that's a decoy!

1 = 7f + 7t
3 = f + t

f = 3-t

1 = 7(3-t) + 7t ---->>> 1 = 14 - 7t +7t ------>>> the variable vanishes! Phew! we have a senseles math 1= 14!
Once your variable vanishes into a senseless math just know that mathematically that equation doesn't exist.

So u have two variables with one equations. We can't solve it!

NOT SUFFICIENT

I call this method the *mathist* method. When u are trying to analyse the question into getting a straight out answer and you couldn't or your mind gets fuzzy, just get down and mathy with the your pen.

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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]
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gmatt1476 wrote:
The first four digits of the six-digit initial password for a shopper's card at a certain grocery store is the customer's birthday in day-month digit form. For example, 15 August corresponds to 1508 and 5 March corresponds to 0503. The 5th digit of the initial password is the units digit of seven times the sum of the first and third digits, and the 6th digit of the initial password is the units digit of three times the sum of the second and fourth digits. What month, and what day of that month, was a customer born whose initial password ends in 16 ?

(1) The customer's initial password begins with 21, and its fourth digit is 1.
(2) The sum of the first and third digits of the customer's initial password is 3, and its second digit is 1.

DS21891.01

A-B-C-D-1-6

Since A-B represents a day between 01 and 31, inclusive:
A = 0, 1, 2, 3

Since C-D represents a month between 01 and 12, inclusive::
C = 0 or 1

The 5th digit of the initial password is the units digit of seven times the sum of the first and third digits:
1 must be the units digit of a multiple of 7, suggesting the following:
7*3 = 21
Implication:
A and C -- the first and third digits -- must SUM TO 3, as follows:
A=2 and C=1
A=3 and C=0

The 6th digit of the initial password is the units digit of three times the sum of the second and fourth digits.
6 must be the units digit of a multiple of 3, suggesting the following:
3*2 = 6
Implication:
B and D -- the second and fourth digits -- must SUM TO 2, as follows:
B=0 and D=2
B=1 and D=1
B=2 and D=0

Statement 1: The customer's initial password begins with 21, and its fourth digit is 1.
First four digits:
2-1-C-1
Since A=2 and A+C=3, C=1.
Result:
2-1-1-1 --> 21 November
SUFFICIENT.

Statement 2: The sum of the first and third digits of the customer's initial password is 3, and its second digit is 1.
A-1-C-D
Since B=1 and B+D=2, D=1.
Result:
A-1-C-1.

If A=2 and C=1, we get:
2-1-1-1 --> 21 November

If A=3 and C=0, we get:
3-1-0-1 --> 31 January

Thus, the date of birth cannot be determined.
INSUFFICIENT.

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The first four digits of the six-digit initial password for a shopper' [#permalink]
By any means this problem has a lengthy/time consuming solution.
Could experts please advise what a test taker should do with such questions in exam - skip or attempt?
GMATGuruNY
chetan2u
Bunuel

Thanks!
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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]

Interesting read about punting. When I feel the need to punt I have started to use the "number of variables approach" (promoted, if I remember correctly, by MathRevolution) as a last resort and my accuracy on these guesses has been very good.

On this question, from statement 1 we are left with only one unknown digit and the info is likely to be sufficient. Statement 2, on the other hand, leaves us with three unknown digits and is less likely to be sufficient.

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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]
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Bambi2021 wrote:
Interesting read about punting. When I feel the need to punt I have started to use the "number of variables approach" (promoted, if I remember correctly, by MathRevolution) as a last resort and my accuracy on these guesses has been very good.

I could show you GMAT books written in the 1990s (that's well before my time in the field, but I've seen the books) that promote a 'count unknowns, count equations' strategy for DS. It's not new, and it was popular long before the company you mention even existed. But most people stopped teaching it (if they ever taught it at all), because it doesn't work, or at least it doesn't work any more. It might have worked back when ETS developed the GMAT (that's quite a while ago now), but when ACT took over, GMAT math got a lot harder, and DS questions started testing a lot more of the exceptions to rules. I've studied how successful the strategy is on large pools of official questions, and while it works moderately well on easy questions (which aren't questions midlevel or better test takers need a guessing strategy for), it works about as well as random guessing on medium and hard questions (it answers only 1/3 of them correctly, which is about as well as you do when you just rule out the answer choices that are obviously wrong, e.g. when a question asks "What is the value of positive integer x?" and Statement 1 says "x is a prime number", you can instantly rule out A and D).

I'd add that, strictly applied, the strategy doesn't even work on this question. You'd certainly correctly conclude Statement 1 is sufficient, but using Statement 2, you only have three unknowns left once the statement tells you a digit, and you appear to have three equations (or three relationships similar enough to equations that I imagine anyone using the strategy would count them as such). So someone using the strategy would presumably pick D, which is the wrong answer. Of course, since this strategy works 1/3 of the time on questions at this difficulty level, you'll find plenty of questions where it does in fact get you the right answer, but they'll be outnumbered by questions where it fails.
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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]
Bambi2021 wrote:

Interesting read about punting. When I feel the need to punt I have started to use the "number of variables approach" (promoted, if I remember correctly, by MathRevolution) as a last resort and my accuracy on these guesses has been very good.

On this question, from statement 1 we are left with only one unknown digit and the info is likely to be sufficient. Statement 2, on the other hand, leaves us with three unknown digits and is less likely to be sufficient.

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Bambi2021 - I am glad that it is working out for you but be mindful that a punt is a punt when you spend no more than 30 secs on it. It brings in value in terms of saved time.
I am on the same page as IanStewart above (which is often the case, I admit!). In my experience, GMAT tests far too many "exceptions" for such "rules/strategies" to make much/any difference.
GMAT rewards strong fundamentals and good understanding. Rest all is questionable.
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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]
IanStewart wrote:
I could show you GMAT books written in the 1990s (that's well before my time in the field, but I've seen the books) that promote a 'count unknowns, count equations' strategy for DS. It's not new, and it was popular long before the company you mention even existed. But most people stopped teaching it (if they ever taught it at all), because it doesn't work, or at least it doesn't work any more. It might have worked back when ETS developed the GMAT (that's quite a while ago now), but when ACT took over, GMAT math got a lot harder, and DS questions started testing a lot more of the exceptions to rules. I've studied how successful the strategy is on large pools of official questions, and while it works moderately well on easy questions (which aren't questions midlevel or better test takers need a guessing strategy for), it works about as well as random guessing on medium and hard questions (it answers only 1/3 of them correctly, which is about as well as you do when you just rule out the answer choices that are obviously wrong, e.g. when a question asks "What is the value of positive integer x?" and Statement 1 says "x is a prime number", you can instantly rule out A and D).

I'd add that, strictly applied, the strategy doesn't even work on this question. You'd certainly correctly conclude Statement 1 is sufficient, but using Statement 2, you only have three unknowns left once the statement tells you a digit, and you appear to have three equations (or three relationships similar enough to equations that I imagine anyone using the strategy would count them as such). So someone using the strategy would presumably pick D, which is the wrong answer. Of course, since this strategy works 1/3 of the time on questions at this difficulty level, you'll find plenty of questions where it does in fact get you the right answer, but they'll be outnumbered by questions where it fails.

Nice call Ian. Thanks for the lesson.

Maybe I should not call this a punting strategy, but more of a last resort guessing strategy. Its not that I apply the strategy strictly, thats true, but more that I consider the number of variables as one of the factors in trying to assess the degree of freedom still left to get rid of. Another factor could be whether any upper limit is given. Its not unlikely Ive already been able to eliminate 2 or 3 options when Im facing such a situation.

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Re: The first four digits of the six-digit initial password for a shopper' [#permalink]
gmatt1476 wrote:
The first four digits of the six-digit initial password for a shopper's card at a certain grocery store is the customer's birthday in day-month digit form. For example, 15 August corresponds to 1508 and 5 March corresponds to 0503. The 5th digit of the initial password is the units digit of seven times the sum of the first and third digits, and the 6th digit of the initial password is the units digit of three times the sum of the second and fourth digits. What month, and what day of that month, was a customer born whose initial password ends in 16 ?

(1) The customer's initial password begins with 21, and its fourth digit is 1.
(2) The sum of the first and third digits of the customer's initial password is 3, and its second digit is 1.

DS21891.01

I will not fret over the wordings, but will take home the point that there are certain relations between digits and last two digits are 16.

(1) The customer's initial password begins with 21, and its fourth digit is 1.
So the password with combined information becomes 21A116.
Thus, one unknown digit A. But there is a relationship involving third digit.

The 5th digit of the initial password is the units digit of seven times the sum of the first and third digits
So 1 is the units digit of 7 times (2+A) => 7(2+A), which has a units digit 1.
Now ONLY 7*3 gives a units digit 1, making 2+A=3……A=1.
Number is 211116, that is 21st of Nov.
Sufficient

(2) The sum of the first and third digits of the customer's initial password is 3, and its second digit is 1.
So the password with combined information becomes B1A116.
Thus, two unknown digits. But there is a relationship involving first and third digit.

The 5th digit of the initial password is the units digit of seven times the sum of the first and third digits
So (B+A) =3 => 7(1+A).
B is first digit of date so it can be 0, 1, 2 or 3.
A is the first digit of month, so 0 or 1.
Thus, (B,A) can be (3,0) or (2,1)
Thus, number can be 310116 that gives date as 31st of Jan.
Or, it can be 211116 that gives date as 21st of Nov.
Both dates are valid, so two possibilities.
Insufficient

A
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