ABC * DEF
× 1 2 3 A D 4 B E 5 C F 6
What is A B C + D E F ?
The answer is 835.
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5 solutions
Haha, that's one way to solve it! Did you enjoy this problem?
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Of course! Can we be friends?
I tried divisors of 123456 it is divisible by 4, 8, 16, 32 then tried multiple of 32 which is 3 digits i.e 128, 160, 192, 224
out of these 160 is surely not as unit digit will be 0 similarly 224 is not as digits 2 are repeated.
then tried 128 quotient is not integer.
so remaining 192 is confirmed. then it is easy to find other factor 643
Yes, we essentially have to prime factorize and them try and find the numbers. It is with a slight stroke of luck that there is only 1 way to write it as the product of 2 3-digit numbers. I wasn't expecting that.
Only {192, 643} or (643, 192} can satisfy.
192 + 643 = 835
You should have written a solution for it. How did you got 192 and 643? There is no proper solution over here
It is quite simple. We just have to factorize the product. 123456=4x4x4x3x643=192x643. So our first number is 192 and second number is 643. I know that I have factorized it in short but you can understand it. And last step is to add them. 192+643=835
This implies that we need to find two 3-digit number such that their product is 1 2 3 4 5 6 . It easy to show that it factors to 2 6 × 3 × 6 4 3 . Because 6 4 3 < 6 7 6 = 2 6 2 , so we just need to verify that 6 4 3 is prime by proving that it's not divisible by any primes less than 2 6 . So one of the number we are looking for is 6 4 3 , and the other is simply 2 6 × 3 = 1 9 2 , thus their sum is 8 3 5
To make the question harder: replace A B C , D E F , 1 2 3 4 5 6 to A B C D E , F G H I J , 1 2 3 4 5 6 7 8 9 respectively.
Oh nice. You can post that version.
I thought it was hard to get a unique pair of 3-digit factors, so I liked this question.
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I think it would be impractical to post that version in Number Theory section because it's ridiculously difficult to find other prime factors other than 3 . But it would be too easy in the Computer Science section.
I tried to solve your question, but it seems to be very impossible. This is my concluded factors for that particular number, {10821,11409} . These two numbers were the only 5 digit number that could ever replace for that digits of ABCDE and FGHIJ, but doesn't satisfy the equation above.
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Why not? They are both five digit numbers.
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because.. if we let A=1 , J=0 , H=8, B=2, D=4 and I=9, this will be the result.
AJHBA x AADJI = 123456789
10821 x 11409 = 123456789 on which, ABCDE x FGHIJ is not satisfied...
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@Keil Cerbito – I was referring to this:
× 1 2 3 4 A F 5 B G 6 C H 7 D I 8 E J 9
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@Pi Han Goh – From your stated equation above. this is what it really looks like if I'm going to use the only two factors satisfying that equation.
\[ \begin{array}{ } &&&&&&1&0&8&2&1\\ \times &&&&&&1&1&4&0&9\\ \hline &&1&2&3&4&5&6&7&8&9\\ \end{array}\]
As you see, there are other numbers repeated like 1's , 0's etc. Then, if we let A=1 , J=0 , H=8, B=2, D=4 and I=9, this will be the result, wherein A is not equal to B , is not equal to C , and so on.
\[ \begin{array}{ } &&&&&&A&J&H&B&A\\ \times &&&&&&A&A&D&J&I\\ \hline &&1&2&3&4&5&6&7&8&9\\ \end{array}\]
Unless, if you define that A can be equal to B or equal to C etc, thus you're just only pertaining to " Find the possible 5-digit factors of 123456789 and that of the digits "0,1,2,3,4,5,6,7,8,9" can be repeated on the factors.
I guess its already quite harder for me. Why do you want to make it more difficult for a 8th standard student.
676 = 26^2 and not 24^2
I just kept dividing by 2 and got 1929. I realised the highest power of 2 123456 is divisible by is 64. Then I divided 1929 by 3 to get 643 which is a prime. The two factors must be (64*3) and 643 or 192 and 643.