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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:

Ques.jpg [ 6.73 KiB | Viewed 5434 times ]

What is the value of D + O + R + F? (A) 17 (B) 20 (C) 22 (D) 23 (E) Multiple values possible

Re: What is the value of D + O + R + F? Question of the Day III [#permalink]

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14 Nov 2010, 07:23

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if we look at the result, D is carried forward, so D must be 1 -- A F and F adds to give unit digit R. That means R is even and either R is 2F or R is 10 + 2F -- B

now as O and I adds up and give unit digit O..C so fron B and C, we know that R is 10 + 2F and I has to be 9

now again R and F multiply to give unit digit F..D

so from B and D, we know that R is 6 and F is 8 >> O = 5

Re: What is the value of D + O + R + F? Question of the Day III [#permalink]

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14 Nov 2010, 17:47

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.
_________________

I logged on a number of times thinking that I will put up the explanation for this question. But every time I did, I stopped on some other interesting looking question and started replying to that... Finally, I have pulled myself away from everything else and am going to provide the solution here first!

First of all, focus on the big picture. IF * R = OFF IF * D = IF ... From here, we can say that D = 1. Nothing else possible. and OFF + IF = DORF

Now, two other interesting things:

1. F + F has unit's digit of R.

2. Also O + I gives O as unit's digit and 1 as ten's digit. e.g. if O = 4, 4 + I = 14. This is possible only when I = 9 and there was a 1 carry over from the previous addition of F + F. Then F must be greater than 5 to have a carry over of 1. It cannot be 5 because then R will be 0 but F * R has unit's digit of F, not R (0).

Another interesting thing: F * R has unit's digit of F. This is possible only when F = 0 or F = 5 or R = 6 (Think of multiplication tables of numbers to convince yourself why this is so) Since F has to be greater than 5 (as seen above), R must be 6. If R = 6, then F + F has unit's digit of 6 so F = 8. When we multiply I by R, we get O in ten's digit. I = 9 and R = 6 so O = 5 D + O + R + F = 1 + 5 + 6 + 8 = 20

This question uses your understanding of numbers and how they are added and multiplied. It certainly takes time to get to the answer. Such questions can help you get a feel for numbers and their behavior.
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16 Jul 2014, 22:06

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Re: What is the value of D + O + R + F? Question of the Day III [#permalink]

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Re: What is the value of D + O + R + F? Question of the Day III [#permalink]

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18 Sep 2015, 10:41

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:

Ques.jpg

What is the value of D + O + R + F? (A) 17 (B) 20 (C) 22 (D) 23 (E) Multiple values possible

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

Re: What is the value of D + O + R + F? Question of the Day III [#permalink]

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26 Sep 2015, 01:18

kunal555 wrote:

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:

Ques.jpg

What is the value of D + O + R + F? (A) 17 (B) 20 (C) 22 (D) 23 (E) Multiple values possible

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.

What is the value of D + O + R + F? Question of the Day III [#permalink]

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01 Apr 2016, 03:06

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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Re: What is the value of D + O + R + F? Question of the Day III [#permalink]

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12 Jul 2017, 03:58

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What is the value of D + O + R + F? Question of the Day III [#permalink]

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20 Jul 2017, 21:16

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.

Re: What is the value of D + O + R + F? Question of the Day III [#permalink]

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09 Aug 2017, 09:12

VeritasPrepKarishma wrote:

I logged on a number of times thinking that I will put up the explanation for this question. But every time I did, I stopped on some other interesting looking question and started replying to that... Finally, I have pulled myself away from everything else and am going to provide the solution here first!

First of all, focus on the big picture. IF * R = OFF IF * D = IF ... From here, we can say that D = 1. Nothing else possible. and OFF + IF = DORF

Now, two other interesting things:

1. F + F has unit's digit of R.

2. Also O + I gives O as unit's digit and 1 as ten's digit. e.g. if O = 4, 4 + I = 14. This is possible only when I = 9 and there was a 1 carry over from the previous addition of F + F. Then F must be greater than 5 to have a carry over of 1. It cannot be 5 because then R will be 0 but F * R has unit's digit of F, not R (0).

Another interesting thing: F * R has unit's digit of F. This is possible only when F = 0 or F = 5 or R = 6 (Think of multiplication tables of numbers to convince yourself why this is so) Since F has to be greater than 5 (as seen above), R must be 6. If R = 6, then F + F has unit's digit of 6 so F = 8. When we multiply I by R, we get O in ten's digit. I = 9 and R = 6 so O = 5 D + O + R + F = 1 + 5 + 6 + 8 = 20

This question uses your understanding of numbers and how they are added and multiplied. It certainly takes time to get to the answer. Such questions can help you get a feel for numbers and their behavior.

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?