What is the value of D + O + R + F? Question of the Day III

took me a while but DORF is 1568 so a total of 20 (B)

the tricky bit is whether I = 4 or 9... only I = 9 works...

I don't understand the second part of that question, after OFF
Does that mean when OFF*IF=DORF ?

or OFF+IF=DORF?

no - the X sign at the bottom is just meant to stand for a 0 --- its like when you multiply, say 45 * 94 the regular way --- when you multiply the 9 in 94 with the 2 digits of 45 you will put a 0 at the extreme right...

That's some pretty good work... The answer is indeed 20. (I thought I had made the question tricky but I guess you guys are really good!)
I will provide the detailed explanation tomorrow.
Don't try to answer it in under two minutes (though if you did, then you are a genius). I made this question with the sole purpose of making you think and bringing some number properties to your notice.

On multiplication of IF with D, we get IF. This means D = 1.
In sum of OFF and IFx O+I gives DO. The unit digit of O+I can be O only if O is added to 10. So, I must be 9 and F+F in the sum should give carry over of 1.
As F+F gives a carry over of 1, F must be greater than or equal to 5.
F*R gives F. This is possible
1) When R is 1
2) When F is 5 and R is 3,7 or 9
3) When F is 8 and R is 6.
As D is already 1, R cannot be 1. Also if F is 5, then sum of F+F in sum of OFF and IFx will give R = 0. This is not possible because if R is 0, Then OFF would be 0.
So, F can only be 8 and hence R can only be 6.
Now, IF is 98 and DR is 16. From multiplication of IF and DR we get O=5
So,
D+O+R+F = 1+5+6+8 = 20

Answer:- B

IMO, this solution is better. we need to follow a proper sequence in order to get the answer, else it becomes time consuming. i got all the values except O, which we can get easily just by multiplication.

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

Tough one and i would not have been able to do it within in 3mins in real GMAT. But the question has given some good insights.
I solved it through brute force

I used following logic

D=1 since D multiplied with IF gives IF

Need a set of Nos (F*R) on whose multiplication we get one's digit as F and also 2F=R

Which are the such nos lets take some examples

In table of 2 we have 2*6=12 now we are getting unit's digit as 2 but when we double F(2) we get 4. Hence not possible.

No such no in table's of 3, 4, 5, 6, & 7

But we get such a no in the table of 8 ie 8*6=48. Units digit is 8 and 2*8=16, the unit digit is 6=R

The equation looks like this now

I 8(F)
1 6(R)
O 8(F) 8(F)
I 8(F)
1 O 6(R) 8(F)

Now we need to plugin other nos. When we multiply 8*6 we get 48 and thus 4 is the carry over. Now as per the question stem tens digit is also same as units digits which means R*I=product +4= should give a 2 digit no with 8 as its units digit.

This no is 9. ie 9*8=54+4=58.

Hence I=9 and O=5

Plugin

I 8(F)
1 6(R)
5(O) 8(F) 8(F)
1 8(F)
1 5(O) 6(R) 8(F)

No is 1568=sum=20

Can you explain the bold part? If F=5 and R=6, then unit of digit is 0 which does not equal 5. And if you plug in any other even number for R when F = 5, then unit digit will be 0 too. Couldn't both F and R be equal to 6?

Hi,
Question is why does F+F would have carry over of 1? How does DO gives us that F+F has carry over of 1. Can't it be F+F could be like 2+2 = 4 and that O+I is 9+5 where O is 9 and I is 5 ?

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.