Wow! Six of Six!

Number Theory Level 2

$\large\begin{array}{lllllll}&&&&\color{#D61F06}A&\color{#20A900}B&\color{#3D99F6}C\\\times&&&&\color{#D61F06}A&\color{#20A900}B&\color{#624F41}D\\\hline&\ \color{#302B94}6&\ \color{#302B94}6&\ \color{#302B94}6&\ \color{#302B94}6&\ \color{#302B94}6&\ \color{#302B94}6\end{array}$

What is the value of $\overline{\color{#D61F06}A\color{#20A900}B\color{#3D99F6}C}+\overline{\color{#D61F06}A\color{#20A900}B\color{#624F41}D}\ ?$

The answer is 1633.

6 solutions

Chew-Seong Cheong
May 8, 2015

Let $\space X = 100A+10B$

$(X+C)(X+D) = 666666 \quad \Rightarrow X^2 \approx 666666$ but $< 666666$

Assuming $\space X = \lfloor \dfrac{\sqrt{666666}}{10} \rfloor \times 10 = 810$

$\Rightarrow (810+C)(810+D) = 656100+810(C+D)+CD = 666666 \\ \Rightarrow C+D = \dfrac {10566-CD}{810} \approx 13 \quad \Rightarrow C=4, D= 9$

Checking the result, $\space 814 \times 819 = 666666$ , which is correct.

Therefore, $\space \overline{ABC} + \overline{ABD} = 814+819 = \boxed{1633}$

Moderator note:

There's a slightly simpler approach to this. Hint: it's almost the square root of 666666, and what two single digit numbers have a last digit of 6 when being multiplied together?

Bonus question: What would the value of $\overline{ABC} + \overline{ADE}$ be if I change the question to this?

$\large\begin{array}{lllllll}&&&&\color{#D61F06}A&\color{#20A900}B&\color{#3D99F6}C\\\times&&&&\color{#D61F06}A&\color{grey}D&\color{#624F41}E\\\hline&\ \color{#302B94}5&\ \color{#302B94}5&\ \color{#302B94}5&\ \color{#302B94}5&\ \color{#302B94}5&\ \color{#302B94}5\end{array}$

Let $C>D$ , $X=\frac{\overline{ABC}+\overline{ABD}}{2}$ and $\Delta = \frac {C-D}{2}$ . Therefore, $\overline{ABC} = X + \Delta$ and $\overline{ABD} = X-\Delta$ .

$\Rightarrow \overline{ABC} \times \overline{ABD} = (X+\Delta)(X-\Delta)=X^2-\Delta^2 = 666666$

$X^2 \approx 666666 \quad \Rightarrow X \approx \sqrt{666666} = 816.4961727 \approx 816.5$

We know that $X$ is half-way between $\overline{ABC}$ and $\overline{ABD}$ and the likely values of $C,D$ for the last digit of their product to be $6$ are $\{ (6,1); (3,2); (9,4)\}$ . Only $(9,4)$ is half-way between $6.5$ . Therefore, $\Delta = \frac {9-4}{2} = 2.5$ and $X = 816.5$ . We note that $816.5^2-2.5^2 = 666666$ indicating the solution is correct. Therefore, $A=8$ , $B=1$ , $C=9$ , $D=4$ and $\overline{ABC} + \overline{ABD} = 819+814 = \boxed{1633}$ .

Chew-Seong Cheong - 6 years, 1 month ago

For $\overline{ABC} \times \overline{ADE} = 555555$ , we know that $A=7$ , since $700^2 < 555555 < 800^2$ . Therefore,

$\overline{ABC} \times \overline{ADE} = (700+\overline{BC})(700+ \overline{DE}) = 555555 \\ \Rightarrow 490000 + 700(\overline{BC} + \overline{DE}) + \overline{BC} \times \overline{DE} = 555555 \\ \quad \space 700(\overline{BC} + \overline{DE}) + \overline{BC} \times \overline{DE} = 65555$

$\Rightarrow 7(\overline{BC} + \overline{DE} )< 655 = 651, 644, 637...$ . Let $\overline{BC} + \overline{DE} = 93-a$ , where $a=0,1,2...$ .

This implies that $\overline{BC} \times \overline{DE} = 100(4+7a) + 55 \quad \Rightarrow \overline{DE} = \dfrac{100(4+7a)+55} {\overline{BC}}$ .

$\Rightarrow \overline{BC} + \dfrac{100(4+7a)+55} {\overline{BC}} = 93-a \\ \quad \space \overline{BC}^2 - (93-a)\overline{BC} + 100(4+7a)+55 = 0$

We note that when $a=1$ , we have:

$\overline{BC}^2 - 92\overline{BC} + 1155 = 0 \\ \Rightarrow (\overline{BC}-15)(\overline{BC}-77)=0\\ \Rightarrow \overline{BC}, \overline{DE} = 15, 77 \\ \Rightarrow \overline{ABC} + \overline{ADE} = 715+777 = \boxed{1492}$

Chew-Seong Cheong - 6 years, 1 month ago

observe that 555555=555(1001)= 37x5x3(7x11x13) now we are looking for a multiple of 37 which lies between 555 and 1001 containing two of the numbers 3,5 7,11and 13 as factors the only number satisfying this requirement is 777(37x7x3) and the rest follows

Des O Carroll - 6 years, 1 month ago

2 and 3. 7 and 8.

Stewart Gordon - 6 years, 1 month ago

Stewart Gordon
May 15, 2015

It's straightfoward to discover that $810 < \sqrt{666666} < 820$ . Once you've worked that out, all you need to do is realise that 666666 is divisible by 11, therefore one of the factors must be. There's only one integer in $[810, 819]$ that's divisible by 11, and so the solution falls into place.

Moderator note:

Quite ingenious. Wonderful!

Vivek Shrivastava
May 12, 2015

Use approximation: 800 x 800=640000. So, A must be 8. Now, 810 x 810 = 656100 and 820 x 820 = 672400. So, B must be 1. Now let's look at the unit number: C and D are not equal and unit number of product of the two numbers results in 6. So, the two numbers can be combination of (2 and 3) or (4 and 9). But, 666666 is closer to 672400 than 656100. Therefore, C and D must be equal to higher values of the two combination i.e. 4 and 9. So, the two numbers must be equal to 814 and 819 and the sum of two numbers must be equal to 1633.

Vishwesh Shrimali
May 10, 2015

Well, I didn't use maths but programming instead. But, here is the program I wrote for solving this question. It can be modified for solving similar problems.

http://ideone.com/QTYU5N

Thanks

Aditya Chauhan
May 9, 2015

666666 is ABC times ABD .

So it is close to a perfect square since first two digits are same.

On checking some values we get A=8 B=1 .

substituting these values we get an equation in C & D.

Solving it v get C,D=4,9 OR 9,4

Therefore ABC + ABD = 1633.

Sorry if you dont understand it, didn't have time and i don't know LATEX.

Moderator note:

Your solution is not clear. What do you mean by "On checking some values we get A=8 B=1 ."? How did you know that? Did you check for all $9000$ possible scenarios?

Challenge Master Note:- Upvoted XD

Mehul Arora - 6 years, 1 month ago

Indra Yudhawan
May 8, 2015

After several calculation, i found that ABC = ABD, so just do 666666^0,5 and then multiply it by 2 _

I got 1632,99 and i answer 1633, dunno that was right ones or just lucky _

Moderator note:

Wrong. If ABC = ABD, then 666666 is a perfect square which is clearly not. And obviously $1632.99 \ne 1633$ .

Wow! Six of Six!

The answer is 1633.

6 solutions

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