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

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

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22 Jun 2017, 13:44
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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
[Reveal] Spoiler: OA

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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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22 Jun 2017, 19:33
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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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24 Jun 2017, 19:40
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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

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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24 Jun 2017, 20:54
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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

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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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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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25 Jun 2017, 17:13
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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08 Aug 2017, 22:08
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(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

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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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15 Aug 2017, 07:39
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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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01 Oct 2017, 16:46
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.

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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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24 Oct 2017, 01:44
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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17 Nov 2017, 11:42
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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Re: In the two-digit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The   [#permalink] 17 Nov 2017, 11:42
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