As simple as ABC

Number Theory Level 1

$\begin{array} { l l l l l } & & & & & 9 & 9 & 9 \\ \times & & & & & A & B & C \\ \hline & & D & E & F & 1 & 3 & 2 \\ \end{array}$

In this cryptogram, $A,B,C,D,E$ and $F$ are (not necessarily distinct) single digits. What is the value of $A+B+C+D+E+F?$

The answer is 43.

5 solutions

Chew-Seong Cheong
May 9, 2015

$999\times \overline{ABC} = \overline{ABC}(1000-1) = \overline{ABC000} -\overline{ABC} = \overline{DEF132} \\ \Rightarrow 1000 - \overline{ABC} = 132 \quad \Rightarrow \overline{ABC} = 868 \quad \Rightarrow 999 \times 868 = 867132 \\ \quad \Rightarrow A+B+C+D+E+F = 8+6+8+8+6+7 = \boxed{43}$

Moderator note:

Good work! Bonus question: What would the answer be if I change the number $999$ to $998$ ?

Applying the same principle, we have:

$998\times \overline{ABC} = \overline{ABC} (1000-2) = \overline{ABC000} - 2\times \overline{ABC}$

$\Rightarrow 2\times \overline{ABC} = \begin{cases} 1000-132 = 868 & \Rightarrow \overline{ABC} = 434 \\ & \Rightarrow 998\times 434 = 433132 \\ 2000-132 = 1868 & \Rightarrow \overline{ABC} = 934 \\ & \Rightarrow 998\times 934 = 932132 \end{cases}$

Chew-Seong Cheong - 6 years, 1 month ago

for 2XABC WHY DID U CONSIDER 2000-132? PLEASE CAN U EXPLAIN?

Bharat Naik - 6 years ago

Because $\overline{ABC}\times 2$ can be larger than $1000$ but $\overline{ABC}$ is definitely less than $1000$ .

Chew-Seong Cheong - 6 years ago

Without having to change DEF132; would you be able to use this principle for 997 × ABC = DEF132?

Thanks in advance

Jakob Johnson - 5 years, 2 months ago

Sir can u plz explain that step of

1000-ABC = 132 how

Sushant Agarwal - 6 years ago

From $\space 999\times \overline{ABC} = \overline{ABC000} - \overline{ABC} = \overline{DEF132}\space$ or

$\begin{array} {} & A & B & C & 0 & 0 & 0 \\ - & & & & A & B & C \\ \hline & D & E & F & 1 & 3 & 2 \end{array} $

Can you get $\overline{ABC}$ from the above? If you do, what you would have done was probably $1000 - \overline{ABC} = 132$ . That is how we normally do in arithmetic. We borrow $1$ in front, because $\overline{000}$ cannot minus $\overline{ABC}$ .

Chew-Seong Cheong - 6 years ago

Yep, he made a mistake there

Luis Rivera - 5 years ago

Awesome solution dude!, greetings from México

Angel T. López - 6 years ago

Even i had got ABC as 868 but then i saw DEF should have been different then ABC but A & D and B & E are equal i.e A=D & B=E. It was not specifically mentioned in the question whether ABC can have values same to values of DEF.

Karthik Thakkar - 6 years ago

999 x ABC = (1000-1) x ABC = ABC000 - ABC = DEF132 .: ABC000 = DEF132 + ABC .: ABC = 868 DEF132 + 868 = 868000 .: DEF132 = 868000 - 868 .: DEF132 = 867132 .: DEF = 867 So, 8+6+8+8+6+7=43

Faraz Ansari - 5 years, 4 months ago

You guys help me with something. If ABC is 868, why where A and C represented with different letters

Uthman Tanko - 2 years, 3 months ago

This method is so short and sweet... My method is so big and complicated... $UPVOTED +1$

Vighnesh Raut - 6 years, 1 month ago

Des O Carroll
May 9, 2015

999x=1000y+132

1000x=x+1000y+132

1000(x-y)=x+132

x+132=1000 and x-y=1

x=868 and y=867

summing the digits of x and y gives the answer=43

Moderator note:

This is correct too!

why do you assume x-y=1 ??

José Isaac Tijerina - 5 years ago

x = ABC => x<1000 => x+132<1132 => 1000(x-y)<1132 => (x-y)<1.132

Since x and y are distinct whole numbers x-y=1

Lishan Aklog - 1 year, 6 months ago

The number is divisible by nine obviously. The Divisibility rules says that if a number is a factor of 9 , all the digits add up to 9. 1 + 3+ 2 = 6. Then it's just a matter. of finding three numbers that add up 21, and when added to the 6, make 27, and 2 + 7 = 9

Luis Rivera - 5 years ago

Vighnesh Raut
May 8, 2015

As the last digit is $2$ , $C$ must be equal to $8$ . As second last digit is $3$ , $9C+7+9B (mod 10)$ must be equal to $3$ . $72+7+9B=79+9B=9+9B=9(B+1) (mod 10)$ . Hence, $B+1=7$ or $B=6$ . Now, solving for A, $999(A)(100)+(999)(68)\equiv 132\quad (mod\quad 1000)\\ (-1)(A)(100)+(-1)(68)\equiv 132\quad (mod\quad 1000)\\ 200+100A\equiv 0\quad (mod\quad 1000)\quad$ so, $A=8$ giving us the $3$ digit number $ABC$ as $868$ .Hence, $999*868=867132$

Moderator note:

There's a simpler solution for this. Hint: $999+1=1000$ .

But this is a solution! And not a bad one at all; being iterative, it’s easy to solve mentally.

Seeing multiple ways to solve a problem is a beautiful thing. That’s why I love reading all the diverse problem solving methods for each question on Brilliant.

A Former Brilliant Member - 3 years, 1 month ago

Lam Nguyen
May 12, 2015

999 = -1 (mod 1000) DEF132 = 132 (mod 1000) => -abc = 132 (mod 1000) <=> abc = 868

Tim Boan
Oct 3, 2017

For 998 vs 999 solution is 21

As simple as ABC

The answer is 43.

5 solutions

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