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rookie1ja

No. 1

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No. 1 - Back to the Cryptograms and Algebra Puzzles

Replace the same characters by the same numerals so that the mathematical operations are correct.

ABCB - DEFC = GAFB		
: + -
DH x AB = IEI
----------------------
GGE + DEBB = DHDG[/code] [size=4][color=#FF0000][b]This old topic is locked since solution is already provided in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers.[/b][/color][/size] [spoiler=Solution][b]No. 1 - solution[/b] A=3, B=8, C=0, D=1, E=4, F=5, G=2, H=7, I=6
[code]3808 - 1450 = 2358
: + -
17 x 38 = 646
-------------------------
224 + 1488 = 1712

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Guest

well I now know the answer to the problem but I would really love to see the solution to this problem and by that I mean I would like to see how the value of the variables was determined without just inputing random numbers.

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Here's how I attempted it:

C=0 is quite obvious from the ***B - ***C = ***B.

From the same summation, it is seen that as C=0, F must be 5 as C carries from B and becomes 10 and then subtract that from 5 gets 5. so F=5.

As 1 was carried from B, so (B-1) - E = A and also A + F = B. So we know from this that F= 1 + E. Hence E = 4.

Then from GAFB - IEI = DHDG, it is clear that A has carried from G so G - 1 = D. Hence G = D + 1 i.e. G = 2.

A - D = G so A = 3 and putting this in A + F = B, you can get B = 8.

I am sure you can take it from here

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Guest

I'm new but I just had to ask...

Quoted from above:

"Then from GAFB - IEI = DHDG, it is clear that A has carried from G so G - 1 = D. Hence G = D + 1 i.e. G = 2."

I'm not sure you can say that with certainty. Where did you deduce D = 1 from? If you got it from

GAFB - IEI = DHDG then you made the assumption that B > I in B - I = G and it did not borrow from F in the next column to make the subtraction. If thats how you thought it through then how did you know that you could safely make that assumption?

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Try with that:

Since ABCB : DH = GGE, and C = 0 and E = 4, B must be an even number.

But B must be greater than 5 (B=5+A), then you have only the 6 and the 8.

If B = 6, from the bottom 4+B=10, then G=0, but C=0, then don't works.

If B = 8, 4+B=12, then G=2.

Best Regards,

Harold

(I appologize for my bad english)

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a and b equals mc2 so you minus d from the entire equation and add pie which gives you a strange decimal. change the decimal to a fraction, add 3/4 and then divide by 7. you sould have the answer, if it's another decimal just round it to the nearest whole number and divide by c. you get c by multiplying d by b and then cross canceling out a which equals a number between one and 20. then you can easily get the final answer

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I found alternate solutions to each one:

For #1: A=8,B=6,C=0,D=3,E=7,F=5,G=4

Thus: 8606+3750=4856

For #2: 52*13=676

For #: 6744+117=6861

My method was a combination of deduction and trial and error.

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if u r Replaceing the same characters by the same numerals then how r u getting dat? if u do den u would get a=1 b=2 c=3 d=4... and so on rite?

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Guest

Whatever happened to simple factoring of polynomials? I must be in over my head. Can I get homework help here?

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This is cool. I am going to make my math teacher do this, it took me 2 hours to figure it out, since 6 AM when I decided to finally*Wink, wink* check the site out... I had to plug every # in every hole. I found out 9 wasn't used at all, LOL

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Maybe I missed something, but couldn't you just set every Letter to zero and say "DONE!!!"

(nothing in the instructions said that a number couldn't be the solution to more than one letter) (Duh)

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Here's how I attempted it:

C=0 is quite obvious from the ***B - ***C = ***B.

From the same summation, it is seen that as C=0, F must be 5 as C carries from B and becomes 10 and then subtract that from 5 gets 5. so F=5.

As 1 was carried from B, so (B-1) - E = A and also A + F = B. So we know from this that F= 1 + E. Hence E = 4.

I don't mean to necropost, but I only understood this up until (B-1) - E = A. How did you get A + F = B and F = 1 + E?

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I don't mean to necropost, but I only understood this up until (B-1) - E = A. How did you get A + F = B and F = 1 + E?

You can deduce that A + F = B from the center column: DEFC + AB = DEBB. Since the hundred's place does not increase when you add A with F, A + F < 10, so A + F must equal B => (A + 5 = B). Using this result and substituting this value for B into: (B - 1) - E = A => ((A + 5) - 1) - E = A, eliminate the A's and you get E = 4.

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