In the number 1@3,52#, @ and #

[canbtg]

In the number 1@3,52#, @ and # represent distinct digits. If 1@3,52# is a multiple of 18, is @ > #?

(1) 5@# is a multiple of 8

(2) 5@# is a multiple of 6

I do not have the OA right now ; I will post it as soon as I get .

Thanks.

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Hi canbtg,

This is a remarkably layered symbolism question (and it's far more difficult than a typical symbolism question on the GMAT). It involves some Number Properties and some basic arithmetic.

We're told that 1@3,52# is a multiple of 18, which means that it's DIVISIBLE by 18. We're asked is @ > #? This is a YES/NO question.

Since that big number is a multiple of 18, it must be EVEN. This means that # MUST be even (so it's limited to 0, 2, 4, 6, 8). Next, since that big number is a multiple of 18, it's also a multiple of 9. Using the 'rule of 9', the digits of that big number must SUM to a total that's divisible by 9.

We have 1+@+3+5+2+# = (11 + @ + #) is a multiple of 9. This means that...

@+# = 7 or 16

This severely limits the possibilities. They are:

@ #
1 6
3 4
5 2
7 0
8 8

Now we deal with the Facts....

Fact 1: 5@# is a multiple of 8

Plugging in the above 5 options, there's ONLY 1 option that's divisible by 8: 552. The answer to the question is YES.
Fact 1 is SUFFICIENT

Fact 2: 5@# is a multiple of 6

Plugging in the above 5 options, more than one option is divisible by 6:
516 and the answer to the question is NO.
552 and the answer to the question is YES.
Fact 2 is INSUFFICIENT

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

[Matt@VeritasPrep]

My approach:

We want to do most of the work BEFORE evaluating the two statements, so let's see what we can extract from the stem.

Since 1@352# is a multiple of 18, it must be a multiple of both 9 and 2.

Any multiple of 2 is even, so # is even. Any multiple of 9 has digits that sum to a multiple of 9, so (1 + @ + 3 + 5 + 2 + #), or (11 + @ + #) is a multiple of 9.

Since @ and # are each at most 9, we know that (11 + @ + #) can only equal 18 or 27. (It can't equal any other multiples of 9, as all of them would require @ + # to be negative or to be greater than 25).

Let's take the easier case first. If 11 + @ + # = 27, and # is even, then we must have @ = 8 and # = 8. (Any other numbers are too small.)

The other case has a few more possibilities. If 11 + @ + # = 18, and # is even, then we can find all the possibilities by trying each even value of #. We have

# = 0, @ = 7
# = 2, @ = 5
# = 4, @ = 3
# = 6, @ = 1

Combining these with the other option we found (# = 8, @ = 8), we have FIVE possibilities in total. Off we go to the statements!

S1 tells us that 5@# is a multiple of 8. Looking at our five options above, we have these possibilities:

570, 552, 534, 516, 588

Noticing that 560 (or 8 * 70) is a multiple of 8 in this range, we can quickly determine that 552 is the only multiple of 8 of the five. Hence 5@# = 552, and @ > #. SUFFICIENT!

S2 tells us that 5@# is a multiple of 6. But ALL FIVE of our numbers are multiples of 6, since all of them are even and all of them divide by 3 (check the sums of the digits). Sometimes @ > # and sometimes it isn't, so this statement is INSUFFICIENT.

Pretty neat Q!