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

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

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12
10
AbdurRakib wrote:
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

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.

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Posts: 8332
Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  [#permalink]

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4
2
AbdurRakib wrote:
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

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

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35
7
(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
##### General Discussion
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Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  [#permalink]

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1
1
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
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Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  [#permalink]

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10
1
2
AbdurRakib wrote:
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

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.
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Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  [#permalink]

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864 = 2*2*2*2*2*3*3*3.
As per the question the only factors possible with tens digit 2_ and 3_ are 24 and 36 or 27 and 32.
Either condition is sufficient.
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Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  [#permalink]

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1
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
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Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  [#permalink]

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Can't this be solved without using any of the given two statements?
The information given in the initial statement is already sufficient to arrive at one unique solution? The following statements are just helping us by easing it down.

Sent from my XT1053 using GMAT Club Forum mobile app
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Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  [#permalink]

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It can be 4 and 6 or 6 and 4, how can you be sure about which one of those is the square..
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Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  [#permalink]

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1
AbdurRakib wrote:
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

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

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

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ScottTargetTestPrep wrote:
AbdurRakib wrote:
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

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.

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 Intern  B
Joined: 23 Aug 2017
Posts: 23
Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  [#permalink]

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1
AbdurRakib wrote:
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

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.
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Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  [#permalink]

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let say a and b are the number presented by square and angle.
a+b=10, but the product has unit number of 4. so, there are the following case
a=6, b=4, or a=4 b=6
now we plug in these number to see one of them make product of 864. that is simple
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Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  [#permalink]

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AbdurRakib wrote:
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

Lets do prime factorization of 864, Once done we get

$$2^5 * 3^3$$

We can get 864 by 36*24 or 32*27, there are other pairs as well, but those pairs don't help us

from 1) only one case satisfies the condition, 36*24

from 2) only one case satisfies the condition, 36*24

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

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1
dave13 wrote:
ScottTargetTestPrep wrote:
AbdurRakib wrote:
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

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.

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 Hi dave13

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

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okay so.. 3a * 2b = 864 where a and b are the units digits of the integers 3a and 2b respectively.

statement 1 : a+b =10 ....now focus on the unit digit of 864 ie 4. We need 4 as the unit digit so only possible choices are 6 & 4 or 4 & 6... ie 36*24 or 34*26... clearly, 36*24 is the only possible case as 34*26=864.. Hence a=6, sufficient.

statement 2 : a*b=24 ... ie again we have two cases 36*24 or 34*26... clearly, 36*24 is the only possible case as 34*26=864.. Hence a=6, sufficient.

Hence both statement 1 and statement 2 are sufficient. In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The   [#permalink] 02 Jun 2019, 08:02
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# In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The  