× B E B E E 5 S 9 5 T 9 1
If each letter represents a different nonzero digit, what must B be?
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Nice! This was my first instinct as well. Times 99 made me think immediately of divisibility rules for 9 and 11.
Thanks! I had never seen your "divisibility-by-11" rule before. If there is anyone out that also has not seen your rule before then this little tidbit might help: BEE551 is a 6-digit number. If BEE551 is a multiple of 11 then it must be the product of 11•(5-digit number). The (5-digit number) may be represented as JKLMN or [J][K][L][M][N] where [J]{K][L][M][N] means N•1 + M•10 + L•100 + K•1000 + J•10000 and J, K, L, M and N are each single digits. 11•JKLMN is equal to JKLMN + JKLMN0.
11•JKLMN = [J][J+K][K+L][L+M][M+N][N} which is an "equivalent" 6-digit number that is a multiple of 11.
The rule can be generalized to any N-digit number. Specifically , for our 6-digit number if you sequentially and alternately add and subtract the digits you get 0: –J + (J+K) – (K+L) + (L+M) – (M+N) + N = 0.
So that's how I understand your "divisibility by 11" rule.
Your rule is a much more generalized version of the "divisibility-by-11" recursively reductive rule that I was familiar with: if Pq represents a number K where it is understood that K = q + P•10 and that q is a single digit whereas P may be any number of digits then Pq is a multiple of 11 if and only if P-q is a multiple of 11.
Here's an example of the recursive reductive "divisibility-by-11" rule:
Is 133551 a multiple of 11? Yes, IFF 13355 – 1 = 13354 is a multiple of 11.
Is 13354 a multiple of 11? Yes, IFF 1335 – 4 = 1331 is a multiple of 11.
Is 1331 a multiple of 11? Yes, IFF 133 – 1 = 132 is a multiple of 11.
Is 132 a multiple of 11? Yes, IFF 13 – 2 = 11 is a multiple of 11.
11 is obviously a multiple of 11 with the result that 133551 is also a multiple of 11.
Another example:
Is 678 a multiple of 11? Yes, IFF 67 – 8 = 59 is a multiple of 11.
Is 59 a multiple of 11? Yes, IFF 5 – 9 = -4 is a multiple of 11.
–4 is obviously not a multiple of 11 with the result that 678 also is not a multiple of 11.
Finally, –6 + 7 –8 = –7 ≠ 0 so 678 is confirmed to not be a multiple of 11.
Again, THANKS!
Your solution is nice! It's a fast way to solve this problem!
Great solution, assuming the problem is solvable, which I guess is okay. If the 99 had been, say 77; then, you would reach the same conclusion; but, there would be no values for E, S, and T that would make it work.
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Thanks for bringing up this point :) I solved it like a test! There is only one correct choice.
Phenomenal and deceptively simple insight, Kazem :)
Nice one! One criticism: you have not proven that the problem actually has a solution. You only have shown that, if there is a solution, then B = 1 .
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Thanks :) Actually, I looked at choices. If there was another choice, I would continue my reasoning to show that indeed there is a solution.
It goes like this: B E E 5 5 1 is divisible by 9 . Then 2 E + 3 is divisible by 9 . Thus E = 3 . Divide 1 3 3 5 5 1 by 9 9 to get other digits.
Holy Nellie... What a great approach! (+1)
@Kazem Sepehrinia , do u mind if i make a problem named "in memory of maryam mirzakhani"?
Since 9 9 ⋅ B E S T = 1 0 0 ⋅ B E S T − B E S T , we rewrite the problem as B B E E B E S E 5 T S 5 0 T 1 0 + This can easily be solved working from right to left. We find 1 3 4 9 + 1 3 3 5 5 1 = 1 3 4 9 0 0
Nice insight, Arjen!
Ah, great approach!
Great! I used the same method! Is easier to solve a sum than a multiplication!
Good solution. I did this a long way around figuring out first T then S then trying to find E
which I only did by trial and error, which I know is a totally unsatisfactory way of doing the problem. Good thinking by you! Regards, David
B E S T * 99 can be written as : BEST*(100-1) = BEST00-BEST as shown in the following :
B E S T 0 0
_ B E S T
B E E 5 5 1
T = 9
S = 4
E = 3
It is obvious that there is no carry involved in the subtraction involving B ( since the leftmost digits BE are brought into the final total as they are)
Hence, S-B = E
B = S - E
= 4 - 3
= 1
Same method:)
I mixed the methods, I guess - you can see T = 9 from the original sum, and put that into BEEST00 - BEST
I did that too
Take your pick.
Thank you for restoring the upload dialog. I was unable to drop or paste into the "drop pictures here" form.
9T = 81, T = 9 9S + 8 + 1 = 45, 9S = 36, S = 4 9E + 4 + 4 = 35 E = 3
9B + 3 + 1 = 13, B = 1
One can actually check that is the solution.
Easiest to convert this to another addition problem by using
then working from right to left easy to get:
1 3 3 5 5 1 1 3 4 9 1 3 4 9 0 0
B E S T × 9 9 = B E E 5 5 1
so, the answer is divisible by 99 or 33 or 11......................[it works everywhere]
so, it could be, .....(1)...... 9 − 4 5 + 4 5 − 9 E + 9 E − 9 B or, 9 B = 9 or, B = 1 ........................[for 99]
secondly, 3 − 1 5 + 1 5 − 3 E + 3 E − 3 B or, 3 B = 3 0r, B = 1 ....................................[for 33]
same for, 11
The problem can be cast as : BEST*(100-1) = BEST00 - BEST which is a much easier subtraction problem since 3 digits are already known
As there are two carry digits and you are only adding two numbers , therefore the maximum number B can be is 1
If you would like to use a bit of brute force...
It's clear that s=4 and t=9 if we consider that BEST x 100 must end in 00. Since BEE551 must be divisible by 9 and 5+5+1=11, we know that B+E+E equals either 7 or 16. Given that each letter represents a different non-zero digit, most possible cases are eliminated.
If B+E+E=7, we could have either B=7 with E=1 or B=5 with E=2. If B+E+E=16, we could have B=6 with E=5 or B=2 with E=7.
Plugging in these values makes the answer easy to verify.
B=1. Here is a program in JustBASIC:
for b=1 to 5
for e=1 to 9:if e=b then [ne]
for s=1 to 9:if s=b or s=e then [ns]
for t=1 to 9:if t=b or t=e or t=s then [nt]
best=1000*b+100*e+10*s+t
bee551=100000*b+10000*e+1000*e+551
if best*99=bee551 then print best;" x 99 = ";bee551
[nt]next t
[ns]next s
[ne]next e
next b
1349 x 99 = 133551.
WOW HOW DID HE/YOU DO IT!
The solution is 133551, so the answer is 1
I found this by using every combination of (B, E, S, T), then minimizing the squared error of BEST * 99 and BEE551. Per the below R code:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 |
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 |
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BEST x 99 = BEST x (100 - 1) = BEST00 - BEST = BEE551
That mean's 10 - T = 1. Thus, T = 9. 9 - S = 5. Thus. S = 4. 8 - E = 5. Thus E = 3. S - B = E or 4 - B = 3. Thus, B = 1 .
As we multiply by 99 by x, the last two digits are 100-x.
100-x= 51 . so, x= 49. S=4, T=9.
Since, green star is 7(+4 in hand), so we must multiply 99 by 3 . So, E=3. So, our product is 33551. To get this 3 in second digit form left, we need a 2 similarly. But this 2 should come with a 3 in hand. So B= ( 2-3)/9= *9/9 = *1. Since, the number is one digit only, so *=0. SO, B MUST be 1.
You can solve it absolutely straightforward
Here is a Python script:
http://ideone.com/3owbYW
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B E E 5 5 1 is divisible by 1 1 . It follows that − B + E − E + 5 − 5 + 1 = − B + 1 is divisible by 1 1 . Thus B = 1 .