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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.
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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


Given that A x B < 10, the possible values of AB are 23, 32, 24 and 42.

It is also given that A * B = C and units digit of (A* A + C * B) = B

Since, A = 4 and B =2 only satisfies above condition, hence AB =42.

Answer: (E)
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irda
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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irda
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

Check other Addition/Subtraction/Multiplication Tables problems from our Special Questions Directory
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irda
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

HI,
I would solve it following way--

first A*B<10 means choice C is out ..
Also from multiplication table above A*B=C...
so looking at the choices digit C= 2*3=6 or 2*4=8..

Checking for the HUNDREDS digit in DEBC..
UNITS digit of A*A + B*C=B..
say C is 6, means either A or B is odd..

1)say A is ODD and B is EVEN....
so A*A + B*C=B..means O*O+E*E=O, but it should be same as B, which is E, so out

2)say A is E and B is O..
so A*A + B*C=B..means E*E+O*E=E, but it should be same as B, which is O, so out..

what rae we left with -- C as 8
so AB =24 or 42..

now we know thereare two possiblitie
1) A=2, B=4, C=8..
A*A + B*C=B
2*2+4*8=4+32=36...
but we should get units digit as B or 6.. not correct

2) A=4, B=2, C=8..
A*A + B*C=B
4*4+2*8=32=...
we get units digit as B or 2.. so correct

ans 42
E
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AB
x CA
---------
DEBC

A*B<10

Case 1: A = 2
From the available options satisfying the condition B = {3,4}
Case 1a : B = 3
AB = 23
CA = C2
C = 6;
23*62 = 1426 ; B = 3 and not 2 ; Not possible
Case 1b: B = 4
24*C2 = DE48; C = 8
24*82= 1968; B = 4 and not 6; Not possible

Case 2: A = 3
From the available options satisfying the condition B = 2
AB = 32
AB*CA = 32*C3 = DE2C; C = 6
32*62 = 1984; But B = 2; Not possible

Case 3: A = 4; B = 2
AB*CA = 42*C4 = DE28; C = 8;
42*84 = 3528; D = 3; E = 5; The solution satisfies all condition and is feasible
AB = 42

IMO E
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