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In the twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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22 Jun 2017, 13:44
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In the twodigit 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 twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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17 Nov 2017, 11:42
AbdurRakib wrote: In the twodigit 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 30something number and a 20something 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
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Re: In the twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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08 Aug 2017, 22:08
(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 twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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22 Jun 2017, 19:33
AD BCE From the question stem we know that ◻ and △ are single, nonnegative 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 BCE2) 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 AChoice 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 twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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24 Jun 2017, 19:40
AbdurRakib wrote: In the twodigit 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 twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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24 Jun 2017, 20:54
AbdurRakib wrote: In the twodigit 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 twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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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 twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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15 Aug 2017, 07:39
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 twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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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. Sent from my XT1053 using GMAT Club Forum mobile app



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Re: In the twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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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 twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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14 Jun 2018, 01:23
AbdurRakib wrote: In the twodigit 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+600864=0 or, 20x+30y+xy264=0..................... (Equation 1)Now, 1)The sum of \(\Box\) and \(\bigtriangleup\) is 10. x+y=10 y=10x the above (Equation 1) will become two degree solvable equation. 20x+ 30(10x)+x (10x)264=0 20x+30030x+10xx^2264=0 x^2=36 x=+6 or 6 (ve can not be possible as per question stem). Sufficient2) 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) SufficientAnswer: D



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In the twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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17 Aug 2018, 03:19
ScottTargetTestPrep wrote: AbdurRakib wrote: In the twodigit 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 30something number and a 20something 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 Dear ScottTargetTestPrepGreetings 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



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Re: In the twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The
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11 Oct 2018, 04:50
AbdurRakib wrote: In the twodigit 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.




Re: In the twodigit integers 3[m]\Box[/m] and 2[m]\bigtriangleup[/m],The &nbs
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