VeritasPrepKarishma wrote:
Not too many question today so I thought I will put one up for you to try. It's an alphametic - a mathematical puzzle where every letter stands for a digit from 0 - 9. The mapping of letters to numbers is one-to-one; that is, the same letter always stands for the same digit, and the same digit is always represented by the same letter.
The following alphametic shows multiplication of two numbers IF and DR. The product you obtain is DORF. Try and figure out which letter stands in place of which digit.
Attachment:
The attachment Ques.jpg is no longer available
What is the value of D + O + R + F?
(A) 17
(B) 20
(C) 22
(D) 23
(E) Multiple values possible
Flow of thoughts toward solution :
From the problem's statement and common logic : even if OFF and IFx the biggest 3-digits numbers, D cannot be more than 1 => D e {0,1} => because D != 0 => D = 1 (1)
From the problem's statement : F * R => gives an unit digit F ; In other words, F * R = integer*10 + F (1.3)
From (1.3) : R !=0 and F !=0 (1.4)
From the problem's statement and common logic similar to (1) : F + F = 10 + R (gives an unit digit R) = 2F>10 (2)
We should know that : even + even = even ; even + odd = odd ; even * even = even ; even * odd = even (2.0)
(1) + (2) + (2.0) + (1.4) : F - even/odd =>F e {6,7,8,9} (2.1)
(1) + (2) + (2.0) : R - even and !=0 => R e (2,4,6,8) (2.2)
(2.1) + (2.2 ) + (1.3) + (2.0) : R(even) * F(odd/even) = Even => Any even number has unit digit also even => F in even => F e {6,8 ) (2.21)
From (2.2) + (1.3) + (2.21) : We know that R * F = INT*10 + F. Let's start making multiplication : 2*6 != INT*10 + 6 - unsuitable; 2*8 != INT*10 + 8 ... and etc
... Let's apply intuition, logic and common sense => 6 * 6 = INT*10 + 6 - suitable ... 6 * 8 = INT*10 + 8 - suitable and etc ( you can check in excel that indeed only 6 and 8 suitable for this logic ) (2.3)
From multiplication table and (1.3) + (1) + (2) + (2.3) => F e {6,8 } R e {6,8 } + common logic => F = 8 ; R = 6 (2.4)
So we got : IF * DR = (10*I + 8) * 16 = 1"O"68 => 1000 < 1"O"68 < 1999 => 1000 < (10*I + 8) * 16 < 1999 (3)
From (3) : 1000 / 16 < (10*I + 8) < 1999/16 => I e { 6, 7, 8, 9 } (3.1)
From (2.4) + (3) + (3.1) => I != 8 & I !=6 => I e {7,9}
From (2.4 ) + (3) + 78*16 = 1248 => I != 7 => I = 9
Checking : 98 * 16 = 1568
1+5+6 + 8 = 20
Answer is B
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