In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The

Last digit of 864 is 4. Now we need to multiply the last 2 digits of these 2 numbers.
From 1 : sum is 10. Only 4 and 6 satisfies. Sufficient
From 2 : last 2 digits could be 3,8 and 4,6. Now making 23 and 38 or 28 and 33 does not end up to 864. So 4,6 left. Sufficient.

hence D.

Hi,

Various ways..
One as also given above is that 864 is MULTIPLE of just 2 and 3..
So what numbers in 30s are multiple of ONLY 2 and 3...
32 and 36...
1) 32..
Now 2 in 32 requires a 2 or 7 in units digit of 2∆..
But both are different so numbers are 32&27 and their product is 864...
2)36..
6 requires 4 or 9...
24 is MULTIPLE of 2s and 3s.. so ok
But 29 is not so out
Numbers are 36 and 24 and again product is 864..
So possible values 32*27 or 36*24

Now let's see the statements.
I) Sum of unit's digit is 10..
2+7 is 9 so 32*27 is out
6+4=10
Ans is 36*24
Suff
II) product of units digit is 24..
2*7=14.. 32*27 out
6*4=24
Ans is 36*24
Sufficient

D

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AD
BCE

From the question stem we know that ◻ and △ are single, non-negative integers.
We also see that 864 = (\(3^3\)) * (\(2^5\)), so whatever '3◻' and '2△' are they must have prime factors of only 2 and 3.

1) The sum of ◻ and △ is 10 - our symbols can be pairs of: ( {1, 9}, {2, 8}, {3, 7}, {4, 6}, {5, 5} )

Testing
31 and 29: 31 is prime, this can't be an option
39 and 21: 21 = 3 * 7, thus this can't be an option
32 and 28: 28 = \(2^2\) * 7, thus this can't be an option
38 and 22: 22 = 2 * 11, thus this can't be an option
33 and 27: 33 = 3 * 11, thus this can't be an option
37 and 23: 37 is prime, this can't be an option
34 and 26: 26 = 2 * 13, this can't be an option
36 and 24: both numbers end up with only 2 and 3 as prime factors, option valid
35 and 25: 35 = 5 * 7, this can't be an option

Combining the information from the question stem with the information from Statement 1 leaves us with 1 option - Sufficient BCE

2) The product of ◻ and △ is 24 - our symbols can be pairs of: ( {3, 8} and {4, 6} )

Testing
33 and 28: 33 = 3 * 11, thus this can't be an option
38 and 23: 23 is prime, this can't be an option
34 and 26: 26 = 2 * 13, thus this can't be an option
36 and 24: both numbers end up with only 2 and 3 as prime factors, option valid

Combining the information from the question stem with the information from Statement 2 leaves us with one option - Sufficient A

Choice D
-------
This took me far too long though to answer though.
Is there an easier means of testing our various number combinations like this? In hindsight I likely should've written down the numbers from 20 to 29 and 30 to 39 which prime factors were only 2 and 3, and then compared those options with each statement from there:

{24, 27, 32, 36} vs Statement 1
{24, 27, 32, 36} vs Statement 2

we know multiplication ends in 4 so options of last digits : (3,8) , (1,4) ,(8,8) ,(6,4)
statement 1: sum 10 so only 6,4 possible sufficient
statement 2 : product 24 so 3,8 and 6,4 but we know product 864 so check the values : only 6,4 gives the answer
hence both are sufficient therefore D

Let \(\Box\) be x and \(\bigtriangleup\) be y.
Hence, 3\(\Box\)= 30+x & 2\(\bigtriangleup\)= 20+y.
As given (30+x)(20+y)=864
or, 20x+30y+xy+600-864=0
or, 20x+30y+xy-264=0.....................(Equation 1)

Now,
1)The sum of \(\Box\) and \(\bigtriangleup\) is 10.
x+y=10
y=10-x
the above (Equation 1) will become two degree solvable equation.
20x+ 30(10-x)+x (10-x)-264=0
20x+300-30x+10x-x^2-264=0
x^2=36
x=+6 or -6 (-ve can not be possible as per question stem).
Sufficient

2) The product of \(\Box\) and \(\bigtriangleup\) is 24[/quote]
x*y=24
y=24/x
the above (Equation 1) will become two degree solvable equation.
20x+ 30(24/x)+x (24/x)-264=0
x^2 -12x +36=0
x=+6 or -6 (-ve can not be possible as per question stem)
Sufficient


Answer: D

Dear ScottTargetTestPrep

Greetings from a math passionate gmatclub user !

Nice explanation, just one question I have

Okay we made a prime factorization its clear so far

864 = 8 x 108 = 2^3 x 9 x 12 = 2^3 x 3^2 x 2^2 x 3 = 2^5 x 3^3

We see that 2^5 = 32 and 3^3 = 27; so, △ could be 2 and ◻ could be 7.

But suddenly you take a dramatic unpredictable turn,... its like using handbrake in Ferrari you are pro and I am just an amateur driver, beginner so you say "However, we can have other pairs of numbers besides 32 x 27 that multiply to be 864.

For example, one number could be 2^2 x 3^2 = 4 x 9 = 36 and the other number could be 2^3 x 3^1 = 8 x 3 = 24.
"

But how can I figure it out (another combination) so quickly under time constraint ? I would appreciate if you could get the chance to explain it

thank you and have a nice weekend

Dave

I think I found a way to solve without any calculation:

We may say that 3\(\Box\) equals 30 + \(\Box\) and 2\(\bigtriangleup\) equals 20 + \(\bigtriangleup\), then we have this equation (30 + \(\Box\)) * (20 + \(\bigtriangleup\)) = 864, in other words, we have one equation with two variables. We need just one more not equivalent equation with these two variables to find \(\Box\).

Each of the statements give me a not equivalent equation with the variables \(\bigtriangleup\) and \(\Box\), then each of them are sufficient. Option D.

Hi dave13

I dont know if your question is answered.
But here is what i would do .

864=\(2^{5}3^{3}\)

Since one number is in its 20's So ( 20, 21,22,23,25,26,28, 29) cannot be the numbers 23 and 29 are prime and 864 does not have any other prime as its factors except 2 & 3. On the samelines we can reject 20 =5*4, 21=7*3, 22=11*2 and so on

So we are left with 24 and 27

Similarly numbers in 30's we can reject ( 30, 31,33,34,35,37,38, 39) so we are left with 32 and 36

OK
Then we have two statements and we can figure out which combination is true.

Hope this helps

chetan2u

Below was my approach, I do not think we need to find the exact value of \(\Box\), we just need to find whether it's possible to get the value of \(\Box\).

So I wrote all the combinations which will get me a 4 in units place (1,4), (4,1), (2,2), (4,6), (6,4).. so on

Now statement 1 says,

1) \(\Box\) + \(\bigtriangleup\) = 10

so that means it will be either (4,6) or (6,4), but I do not really need to calculate further as it will be either of them and that is enough for me.

Sufficient.

Statement 2 says,

2) \(\Box\) x \(\bigtriangleup\) = 24

so that leaves us with (4,6) or (6,4) and (3,8) or (8,3), but again we know that if both statements are to be answers for a DS question, the value calculated from both statements should be equal and we can't have multiplication of 2 different set of numbers leading to the same result (864). So the pair would be (4,6) or (6,4), and again we do not need to know if it will be 6 or 4, just that we will get a unique value for \(\Box\).

Sufficient.


Can you please help me with the above?

Thanks
Saurabh

Hi,
Yes, you are correct on
We just require to know if we can get a unique value, but it has to be UNIQUE.

You are not correct on two points...
1) We require to know the value of say #, and you have got 6 and 4 for #. But we require to know a UNIQUE value.
2) you cannot take the information from statement 1 to statement 2. so the statement 2 tells you that # can be 4, 6, 3 or 8, so 4 possible values.

Pre-Thinking: We are trying to find the value of \(\Box\)

There's quite a simple, out of the box way to solve this question if you know your divisibility rules.
When you multiply those numbers, you get 864; 864 is divisible by 8.
So you need to determine numbers in the 20's and 30's that when multiplied can produce a number divisible by 8. Either one of the numbers is divisible by 8, or one of the numbers is a multiple of 4 and the other number has to be even..
Also, \(\Box\)*\(\bigtriangleup\) = has a unit digit of 4

1.) \(\Box\) + \(\bigtriangleup\) = 10
The only pair that add up to 10 that produces a units digit of 4 when multiplied is 4 and 6
\(\Box\) can't be equal to 6, because you'll get 26*34, and the product is not divisible by 8, so \(\Box\) HAS to be 4
Sufficient

(2) The product of \(\Box\) and \(\bigtriangleup\) is 24[/quote]
Only values that can count is pair 8 and 3, and 6 and 4.
You can throw out pair 8 and 3 because if you plug those in, neither value gives a product that is divisible by 8.
Only pair 6 and 4 works, and 4 has to be the value of \(\Box\)
Sufficient

Answer: Option D

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Bunuel It was easy to find out 36 & 24 that satisfy both the statements. But how do we ensure that this is the only pair that satisfies each equation? Please help!

3□ and 2△ are two-digit integers where the symbols □ and △ represent different digits. This means they can take any value between 0 to 9.

The product of 3□ and 2△ is equal to 864. This is where a process like prime factorization comes in very handy.

When 864 is prime factorized, we see that 864 = \(2^5 * 3^3\). Therefore, 864 can be written as 32 * 27 or 36 * 24.
This means, □ can be 2 or 6. We need the information given in the statements to determine this uniquely.

From statement I alone, the sum of the symbols = 10. Since 2 + 7 does not give us 10, it has to be 6 + 4. Therefore, □ = 6.

Statement I alone is sufficient to answer the question. Answer options B, C and E can be eliminated, possible answer options are A or D.

From statement II alone, the product of the symbols = 24. Therefore, □ = 6 and △ = 4.
Statement II alone is sufficient. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!
Aravind B T.

** getting square and triangle is too much effort, I'm lazy. I'm using x and y instead of square and triangle**

IMO, the most efficient way is this
(3x)(2y) = 864
3x and 2y are some two-digit number, whose product is 864 --- you can write these number in the form:

\((30+x)* (20+y)= 864\)
\(600+30y+20x+xy= 864\)

\(30y+20x+xy=264\)

now, comes a bit tedious process. we need a combination of digit x & y which when multiplied by 30 and 20 and to each other (xy) gives 264.

\(30y+20x+xy=264\)

no easy way out, gotta plug in them values. in both the statement 1 and 2, you will realise x being 4 and y being 6 gives us the result 264.

D.

cheers.

Hi Scott,

However, there will not be any other pairs of numbers besides 32 x 27 and 36 x 24 that multiply to 864. So, if 864 is a product of a 30-something number and a 20-something number, it must be 32 x 27 (with △ = 2 and ◻ = 7) or 36 x 24 (△ = 6 and ◻ = 4).""

given that this indicates 2 values for △ , how is the answer D?
thanks!