Find the missing number

Number Theory Level 3

$\large 3039162537\square6$

For some integer $n$ , the value of the $(n(n+1)(n+2))^2$ equal to the 12-digit integer shown above with the second last digit omitted. What is the value of this digit?

3 9 0 6

2 solutions

Chan Lye Lee
Nov 5, 2015

$(n (n+1) (n+2))^2$ is a multiple of 9 and so # is either 0 or 9. Next, $(n (n+1) (n+2))^2$ is multiple of 4 and so #6 is a multiple of 4. Hence # $=9$ .

Sorry, but can you elaborate without using # tag ?

Vishal Yadav - 5 years, 7 months ago

He used "#" just to indicate the missing digit. You may call it $x$ if you like, the reasoning is the same, i.e., product of three consecutive integers is always divisible by $3!$ , i.e., by both $2$ and $3$ , hence its square is divisible by both $2^2=4$ and $3^2=9$ . Use the divisibility rule of $9$ to get that the missing digit is possibly $0$ or $9$ . Then, use the divisibility rule of $4$ to conclude that the missing digit must be $9$ , hence $\#=9$ in the number $\overline{3039162537\#6}$ .

In case it isn't explicitly stated that there exists an integer $n$ satisfying the criterions, then you'd have to verify it manually using some approximations and a calculator, but that isn't the case here since the problem guarantees the existence of such an $n$ beforehand.

Prasun Biswas - 5 years, 7 months ago

Not a good explanation

Akhil Krishna - 5 years, 7 months ago

n(n+1)(n+2) is a multiple of 6 for every integer n. So its square is a multiple of 36 and hence the number is a multiple of 9 and 4. We note that the digit sum is 45 so the missing digit must be 0 or 9 because of a number is divisible by 9 its digit sum should also be divisible by 9. And the number is divisible by 4 so it's last 2 digits are divisible by 4 and the last 2 digits can be 06 and 96. We see that 06 is not divisible by 4 so last two digits are 96 and hence the missing digit is 9.

Kushagra Sahni - 5 years, 7 months ago

Jason Martin
Nov 11, 2015

Set this number equal to $N$ . Then $N \approx 3\times 10^{11}$ . Since $N=(n(n+1)(n+2))^2$ , we know $n^6 \le N \le (n+2)^6$ . Then $n \le \sqrt[6]{N} \le n+2$ , or evaluated (rounded to the nearest tenth), $79.8 \le n \le 81.8$ . Thus, $n=80$ or $81$ . If $n$ were divisible by $5$ , then clearly $N$ will be divisible by $100$ , meaning its last two digits would be $00$ , which is false. Thus, $5$ does not divide $n$ , so $n=81$ . Finally, $(81 \cdot 82 \cdot 83)^2 \equiv (86)^2 \equiv 96 \mod 100$ , so the missing digit is $9$ .

Find the missing number

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