challenging equation? (: help!?

Question

A * [(B + C)^(D - E) - F^(G*H) ] / J = 10

Knowing that each variable is a unique, single-digit, nonzero number, and that C - B = 1, and H - G = 3, what is the number ABCDEFGHJ, where each letter is a digit? For example, if A = 2, B = 6, and C = 1, ABC would equal 261.

Answer 1

nahh. u just need to make a lot of assumptions.

Watch this space for answers. I'm working on it.

EDIT: (30mins after posting of question)

I GOT IT !!! XD XD XD

ABCDEFGHJ = 256973148
A = 2
B = 5
C = 6
D = 9
E = 7
F = 3
G = 1
H = 4
J = 8

Checked and verified. =)


EDIT 2: I do have a working, but unfortunately i threw it away upon getting the answer. so i shall try to recall wat i did.

anyway, i did a lot of listing and assuming.

My first step was listing possible values of G and H:
G 1 2 3 4 5 6 7
H 4 5 6 7 8 9 10
G*H 4 10 18 28 40 54 70

I thought power 10 is abit too large, as 2^10 is already 1024. so my very first assumption was G = 1, H = 4.

Then, i realized that B+C must be an odd number.
B 2 5 6 7 8
C 3 6 7 8 9
B+C 5 11 13 15 17

Next, i rearranged the equation:
(B+C)^(D-E) - F^4 = 10 (J/A)
(B+C)^(D-E) - 10(J/A) = F^4

Then, i assumed once again that F < 5, as 5^4 is 3125, quite a large number. so F is either 2 or 3.


(B+C)^(D-E) - 10(J/A) = 16
or
(B+C)^(D-E) - 10(J/A) = 81

as above, B+C is quite a large number to square/cube, as B and C cannot be 2 and 3 respectively (because of F), so i made yet another assumption: D-E is 4 or less.

[well i can't remember wat exactly i did from here on, partly coz it's late and i'm tired.]

Answer 2

ABCDEFGHJ = 256973148


Firstly...
C - B = 1.......C = B + 1
H - G = 3.......H = G + 3

The equation can be simplified as:

A * [ (2B + 1)^(D - E) - F^(G(G+3)) ] / J = 10

1. The difference between the numbers within the square brackets cannot be more than 100 since dividing by J, which is a single digit will not give 10 (assuming that A was 1). Therefore both numbers are supposed to be weighted similarly.

2. Consider the value of 'A'. If the difference were even, then it must be multiplied by 5 before dividing in order to get 10. If it were 5, then 'A/J' has to scale it down accordingly. If it were odd, there are no possible values 'A' can take. Thus the difference can be either be even or a multiple of 5.

3. I assume F is an odd number! The reasoning behind this is that the difference between the 2 numbers can be even [Exception is if the difference were an odd multiple of 5. If it doesn't work then consider the other alternative]. Since (2B + 1) raised to any power is odd, then F must be odd to obtain an even difference.

4. Look at all powers that have values in the range of 1 - 256, since I assume that the values are small and close in value. I consider 1^2 to 15^2, 1^3 to 6^3, 1^4 to 4^4 and 1^5 to 3^5. Look at those with odd bases [also easier to start off with since trial and error must take place now].

5. I obtain and test 11^2 - 3^4 = 121 - 81 = 40. From this, (2B+1) = 11-->B = 5, D-E=2, F = 3 and G^2 + 3G = 4 --> G=1

using 40...A = 2 and J = 8 in order to get a solution of 10

If G = 1 then H = 4 from original equation
If B = 5 then C = 6 from original equation

The remaining numbers are 7 and 9, which allows D = 9 and E = 7 to satisfy the final equation of unknowns [D-E=2]

Figured it out 21 minutes after posting but took me ages to logically decipher steps into words lol.

I hope you understand and it helps!

Great question by the way!

Answer 3

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EDIT: (19 minutes after posting of question)

Way to go Woonie!!!
How did you do that?

How did you decide that C = 6 and B =1?
C - B = 1
6 - 5 = 1
1 = 1 TRUE, but how? and why?

Similarly, how did you decide that H = 4 and G =1?
H - G = 3
4 - 1 = 3
3 = 3 TRUE, but how and why did you chose these values for H & G?

I look forward to your explaination.
Thanx. :o)
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ORIGINAL POST:
Someone else posted this same question about 4 hours ago.

You typed it much clearer. Thank you.

So, this is as far as I got, until I got stumped.

I will ask some fellow colleges at work about this challenge question. It's a good one!!!

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Solution, so far:
A * [(B + C)^(D - E) - F^(G*H)] / J = 10
(J){ A * [(B + C)^(D - E) - F^(G*H)] / J } = (J)(10)
A * [(B + C)^(D - E) - F^(G*H)] = 10J
{ A * [(B + C)^(D - E) - F^(G*H)] } / A = 10J / A
(B + C)^(D - E) - F^(G*H) = 10J / A

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Given #1:
C - B =1
C - B + B =1 + B
C = 1 + B

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Given: #2:
H - G = 3
H - G + G = 3 + G
H = 3 + G

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Substitution:
(B + C)^(D - E) - F^(G*H) = 10J / A
[ B + (1 + B) ]^(D - E) - F^[G(3 + G)] = 10J / A
[ B + 1(1 + B) ]^(D - E) - F^[G(3 + G)] = 10J / A
(B + 1 + B )^(D - E) - F^(3G + G^2) = 10J / A
(2B + 1)^(D - E) - F^(3G + G^2) = 10J / A

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Anyone else care to take it from here. . .?

Good luck!

Keep those challenge questions coming. . . ;o)

Answer 4

You need more knowns before you can solve this.

Answer 5

you must give more information about it