Shaina's five-distinct-digit locker code is 5A48B. What digit letter A symbolizes in Shaina's locker code?

(1) Shaina's locker code is divisible by all integers 2 through 6.

Since the number is divisible by 5 and 2 the B must be 0 (notice that in this case it will also be divisible by 4: a number is divisible by 4 if the last two digits form a number divisible by 4, since 80 is divisible by for so is the number);

The number is also divisible by 3, which means that the sum of the digits must be a multiple of 3: 5+A+4+8+0=17+A ---> A must be 1, or 7. Thus we have two different values of the number. Not sufficient.

(2) Shaina's locker code is divisible by 9 and 11

Divisibility by 9 and 11:
If the sum of the digits is divisible by 9, so is the number --> 5+A+4+8+B=17+A+B=multiple of 9 --> A+B is 1 or 10;

If you sum every second digit and then subtract all other digits and the answer is divisible by 11, then the number is divisible by 11 --> (5+4+B)-(A+8)=1+B-A=multiple of 11 --> B-A=-1 or B-A=10 (not possible, as A and B must be digits and their difference can not be 10);
If A+B=1 and B-A=-1 --> A=1 and B=0;
If A+B=10 and B-A=-1 --> A=5.5 and B=4.5 (not possible as A and B must be digits);

We have only one value of the number: 51480. Sufficient.

Answer: B.
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nice question
(1) says 5A48B is divisible by 2, 3, 4, 5, 6..IF that be so we need to find what does B represents here...since B is divisible by 5 , the value of B can either be 0 or 5...if the value of B is 5, then 5A48B is not divisible by 2,4 and 6 (5A485 is not divisible by 2,4 and 6) thus the value of B is 0 here
so the new term is 5A480..in order for 5A480 to be divisible by 3 or 6( 2 x 3), the total sum of digits should be divisible by 3 so the total makes it 17 + A, now A could be either 1 (total =18) or 4 (total = 21) or 7( total = 24) , thus (1) is not sufficient enough to answer.
lets move to (2) now,
5A48B is divisible by 9 and 11..taking them one by one, to be divisible by 9 , the total should be divisible by 9, the total is 17 + A+B, so A+ B could be either 1( total 18) or 10 ( total 27)
now if the total of A + B is 1, then either A is 0 and B is 1 or A is 1 and B is 0
as we plug them in
the number would be either 50481 or 51480
only 51480 is divisible by 11
so (2) correctly gives the value of A
IMO B
correct me if i am wrong.
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@ Bunuel..
kindly see my explanation above, i have taken a different way..kindly see if it's wrong...regards..
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I think you're missing this part:

If you sum every second digit and then subtract all other digits and the answer is divisible by 11, then the number is divisible by 11 --> (5+4+B)-(A+8)=1+B-A=multiple of 11 --> B-A=-1 or B-A=10 (not possible, as A and B must be digits and their difference can not be 10);
If A+B=1 and B-A=-1 --> A=1 and B=0;
If A+B=10 and B-A=-1 --> A=5.5 and B=4.5 (not possible as A and B must be digits);
From Bunuel.

A+B = 1 ===> 1 or 0, since A and B are positive

A+B = 10 ===> (1 and 9, 2 and 8, 3 or 7, 4 and 6, 5 and 5).

Therefore you can't really assume that the numbers are a and b are 1 and 0. You need 2 equation to solve for 2 variables. Hence the part above gives you another equation, which you use to solve for A and B

rajeev, ur explanation is very simple and i found it easier than Bunuel' explanation. kys123 - even i missed this part of bunuel' explanation....so did the same way as rajeev

The statement I quoted from Bunuel determined if A+B = 1 or A+B=10. Is absolutely needed if the number was larger than 1.

Imagine if the A+B wasn't a easy number like 1. For example if A+B =10 there would be 9 [1n9, 2n8, 3n7, 4n6, 5n5, 6n4, 7n3, 8n2, 9n1] possible ways to arrange the numbers. You wouldn't have time to try all of them. Lets say both of those numbers fit the criteria of being divisible by 11, but there are two pairs of answers; then statement 2 alone would not be able to provide you with sufficient information.

hey dude thanks....got it now. that ws easy i am becoming dumb it seems!!!

I have a question regarding the A part


1) Shaina's locker code is divisible by all integers 2 through 6.
Since the number is divisible by 5 and 2 the B must be 0 (I got this)

From here I used the divisibility rule for 7
(the last digit*2 -other digit) divisible by 7

Thus, 0*2-A-5-4-8= -17-A divisible by 7. Thus -17-A will be -21 and A will be 4

Please let me know if there is anything wrong with my thinking line. Thank you.

We are NOT told that the code number is divisible by 7: "Shaina's locker code is divisible by all integers 2 through 6."
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U did not check for A+B =10 . WHY????? We cud end up with a different value of A. Am I missing something. I think ur solution is not fullproof.

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE


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1. its very simple ... B could only be 0 and sum of all the terms have to me divisible by 3 so it could be 1, 4 or 7. NOT SUFFICIENT

2. 5A48B

for a no. to be divisible by 11 the difference between sum of alternate digits should be 0 .. so 5+4+B=A+8 .. this means A=B+1 ..
now coming to 9, sum should be divisible by 9, 8+4+5=17 .. A+B could be 1 or 11 .. as we know difference between A and B is 1 and none of these could be 10 .. they should be 0 and 1.

I just think if you do the part with 11 first, you can answer it faster or may be I am better with 11s :p

Just one small error..4 is not a possible option as all the digits should be distinct..Does not change the answer though

@ Bunuel I was unable to solve all the 7 questions in 14 mins..Should I be worried if it took me say 18mins or so...

Yes, you are correct. +1.

As for 18 minutes for 7 questions. You mean, these 7 DS questions, right? Those are quite hard, probably 700+, questions, so it's OK to spend more than 2 minutes on each, especially if you got all of them right. Though I'd still advice you to work on speeding up:

HOW TO SPEED UP

Want to speed up? Check this: Timing Strategies on the GMAT
Other discussions dedicated to this issue:
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having-trouble-finishing-quant-section-any-ideas-110613.html
ds-and-timing-help-108310.html
final-weeks-calculation-silly-errors-and-speed-129408.html

Hope this helps.
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Hi Bunuel, is 'If you sum every second digit and then subtract all other digits and the answer is divisible by 11, then the number is divisible by 11' a general rule for multiples of 11?

Yes. The rule is: If you sum every second digit and then subtract all other digits and the answer is divisible by 11, then the number is divisible by 11.
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