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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, 14:44
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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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Md. Abdur Rakib

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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, 20:33
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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, 20:40
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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, 21:54
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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, 18:13
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, 23:08
(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, 08:39
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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] 15 Aug 2017, 08:39
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