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

Question

In the two-digit integers 3 \Box  and 2 \bigtriangleup ,the symbols  \Box  and  \bigtriangleup  represent different digits, and the product (3 \Box )(2 \bigtriangleup ) is equal to 864. What digit does  \Box  represent ?

(1) The sum of  \Box  and  \bigtriangleup  is 10.
(2) The product of  \Box  and  \bigtriangleup  is 24

MuchiMuchi · Accepted Answer

(3◻)(2△) = 864
Write it like this : 
(30 + ◻)(20 + △) = 864
60 + 30△ + 20◻ + ◻△ = 864

Statement 1:  The sum of ◻ and △ is 10.
So ◻ + △ = 10 
So ◻ = 10 - △
Replace in  60 + 30△ + 20◻ + ◻△ = 864  the value ◻ by 10 - △
And then it is a simple equation.

Statement 2:  The product of ◻ and △ is 24
So ◻ and △ are different from 0 
So ◻ = 24/△
Replace in  60 + 30△ + 20◻ + ◻△ = 864  the value ◻ by 24/△
And again it s a simple equation.

So D

ScottTargetTestPrep · Answer

We know the product (3◻)(2△) is equal to 864. Let’s first factor 864:

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.

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. In this case, we have △ = 6 and ◻ = 4.

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).

Statement One Alone:

The sum of ◻ and △ is 10.

This means △ = 6 and ◻ = 4, since the other option has △ + ◻ = 2 + 7 = 9. Statement one alone is sufficient.

Statement Two Alone:

The product of ◻ and △ is 24.

This means △ = 6 and ◻ = 4, since the other option has △ x ◻ = 2 x 7 = 14. Statement two alone is sufficient.

Answer: D

bjp2007 · Answer

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.

chetan2u · Answer

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

Silver65x · Answer

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

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

OC2910 · Answer

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

su30mki · Answer

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 
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

dave13 · Answer

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

devlin · Answer

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.

Probus · Answer

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

Sarjaria84 · Answer

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

chetan2u · Answer

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.

Jsound996 · Answer

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 
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

avigutman · Answer

Video solution from Quant Reasoning starts at 17:30 Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=2QKWyFTyFFU

GMATinsight · Answer

Answer: Option D

Video solution by  GMATinsight

https://www.youtube.com/watch?v=KNUpeG4kFa0

CrackverbalGMAT · Answer

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.

ueh55406 · Answer

** 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.

tickledpink001 · Answer

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!

jabhatta2 · Answer

Hi  avigutman  – this is what I did. I am not sure if I got lucky or was this based on logic

I selected (D) without pen and paper – having gone through your solution – I see pen/paper was used.

Just curious on your thoughts on this method

S2 -- Product of  \Box  and  \bigtriangleup  is 24.

Given 864 is ending with a 4 -- the only pairs for  \Box  and  \bigtriangleup  will be 
(8,3)
(3,8) 
(6,4)
(4,6)

So, then the two numbers would be 
 (28,33) 
(23,38)
(26,34)
(24,36)

THEN   I ASSUMED (THIS WAS A LEAP OF FAITH)

Surely only one pair (From the list in the  blue  above) would be the product of  864 .

Hence (s2) MUST BE sufficient as only pair (From the list in the blue), should give a product of  864

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

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