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In the multiplication above, each letter stands for a different non-zero digit, with A x B < 10. What is the two-digit number AB?
(A) 23 (B) 24 (C) 25 (D) 32 (E) 42
Show your best approach
A two digit number multiplied by another two digit number creating a 4 digit number. if we maximize one number to 99, we would realize that minimum value of another number would be atleast 11. (because 11*99 = 999). Further more A*B < 10 So we can list down the possible numbers. 13, 14, 15, 16, 17, 18, 19 20, 21, 22, 23, 24 30, 31, 32, 33 40, 41, 42 50, 51 60, 61 ...
Now if we look at the options we would realize that only 23, 24, 32, 42 are in the options. So we can eliminate all other cases. Furthermore option C (25) is not in our list, so we can eliminate option C
Here we should understand that when AB is multiplied by CA, units digit is C which is also the value of A*B.
23 :- A = 2, B = 3, C = 6 (23*62) = --26 --------> Tens digit of four digit number (i.e.2) is not the same as B, Eliminate 32 :- A = 3, B = 2, C = 6 (32*63) = --16 --------> Tens digit of four digit number (i.e.1) is not the same as B, Eliminate 24 :- A = 2, B = 4, C = 8 (24*82) = --68 --------> Tens digit of four digit number (i.e.6) is not the same as B, Eliminate 42 :- A = 4, B = 2, C = 8 (42*84) = --28 --------> Tens digit of four digit number (i.e.6) is same as B, Bingo.. GO for Choice E
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Re: In the multiplication above, each letter stands for a differ
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20 Oct 2013, 10:09
1
AB x CA --------- DEBC
The fastest way is to look at answers. 2*5=10 means it doesnt work. We're left with 2*3, 2*4,3*2 and 4*2. Start those kinds of questions from E. 42 * C4 > gives us unit digit (C) of DEBC, which is 2*4=8 (B*A=C). Knowing C we know entire AB*CA = 42*84=3528. Check, if B in DEBC=B in AB. Turns out it is. Check other answers, they will prove that answer is E.
In the multiplication above, each letter stands for a different non-zero digit, with A x B < 10. What is the two-digit number AB?
(A) 23 (B) 24 (C) 25 (D) 32 (E) 42
Show your best approach
A two digit number multiplied by another two digit number creating a 4 digit number. if we maximize one number to 99, we would realize that minimum value of another number would be atleast 11. (because 11*99 = 999). Further more A*B < 10 So we can list down the possible numbers. 13, 14, 15, 16, 17, 18, 19 20, 21, 22, 23, 24 30, 31, 32, 33 40, 41, 42 50, 51 60, 61 ...
Now if we look at the options we would realize that only 23, 24, 32, 42 are in the options. So we can eliminate all other cases. Furthermore option C (25) is not in our list, so we can eliminate option C
Here we should understand that when AB is multiplied by CA, units digit is C which is also the value of A*B.
23 :- A = 2, B = 3, C = 6 (23*62) = --26 --------> Tens digit of four digit number (i.e.2) is not the same as B, Eliminate 32 :- A = 3, B = 2, C = 6 (32*63) = --16 --------> Tens digit of four digit number (i.e.1) is not the same as B, Eliminate 24 :- A = 2, B = 4, C = 8 (24*82) = --68 --------> Tens digit of four digit number (i.e.6) is not the same as B, Eliminate 42 :- A = 4, B = 2, C = 8 (42*84) = --28 --------> Tens digit of four digit number (i.e.6) is same as B, Bingo.. GO for Choice E
How did you know to start plugging in numbers and not try and solve this through theory?
Re: In the multiplication above, each letter stands for a differ
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06 Mar 2016, 20:05
irda wrote:
AB x CA --------- DEBC
In the multiplication above, each letter stands for a different non-zero digit, with A x B < 10. What is the two-digit number AB?
(A) 23 (B) 24 (C) 25 (D) 32 (E) 42
I started first with: 30x30 = 900...so to be 4 digit number, the two numbers must be >30... A - 23*32 - well..definitely less than 900. so out B = 24*X2 = 8 so X must be 8..but if we check further, B is not what we need. so out C = 25*X2 = C=0, but C can't be 0 so out D = 32*X3 = 6, so C=6, but 32*63 = gives different values for B so out only E remains..
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64% (03:00) correct 
