Sequences in Digits

• Let n be the original three-digit number

It is given that the digits in the three-digit number n+2 form a geometric sequence.

And the only possible three digit numbers that form a decreasing geometric progression in its digits are...

• 931

• 842

• 421

• 964

Notice only 842 satisfy the condition where n is a legit three digit number

Hence,

n+2 = 842

n = 840 (notice that the digits do form a decreasing arithmetic progression)

Firstly, we must remember that the digits of the original number form an decreasing arithmetic sequence, where there is a negative common difference between its terms. Let us represent each digit as the following:

• Hundreds digit: $h = a - d$
• Tens digit: $t = a$
• Unit digit: $u = a + d$

Take note that the value of $u$ is increased by $2$ as it is said in the given. Therefore we will have the new value of $a + d +2$ for the units digit. We will only use this new value for the next step, not in all of the process hereafter.

The next step is given that when 2 is added, the digits now form a geometric sequence, in which the product of the first and third term is equal to the square of the second term, that is:

• $hu=t^2$
• $(a - d)(a + d +2) = a^2$

Continuing the solution to find $a$ in terms of $d$ :

• $(a - d)(a + d +2) = a^2$
• $a^2 - d^2 + 2a - 2d = a^2$
• $-d^2 + 2a - 2d = 0$
• $2a = d^2 + 2d$
• $a = (\frac{d^2 + 2d}{2}$ )

We will set this equation aside for the later part of the solution.

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Going back to the equation for the original unit digit, we'll represent $u$ in terms of $d$ , and finally rearrange the whole thing in to a quadratic equation in the form $ax^2 + bx + c = 0$ .

• $u = a + d$
• $u = (\frac{d^2 + 2d}{2}$ + d)
• $u = (\frac{d^2 + 4d}{2}$
• $2u = d^2 + 4d$
• $d^2 + 4d - 2u = 0$

Moving on, we must remember that $d$ must be an integer, because we are talking about digits here. We cannot have digits with a decimal point. So in order to have integer solutions to $d$ , we will use the discriminant of the quadratic formula, which is $b^2 - 4ac$ and we must make it a perfect square.

Substituting the constants into the discriminant:

• $b^2 - 4ac$
• $4^2 - (4)(1)(-2u)$
• $16 + 8u$ ----------> This expression must result to a perfect square.

The next thing I did here is to use the trial and error to put the digits 0 to 9 to $u$ to see if the resulting value is a perfect square. And here are the results:

From the table we can say that when the unit digit is 0 or 6, the discriminant is a perfect square, and we can come up with an integral value of $d$ for each.

The next step is to substitute the desired values of $u$ , which are 0 and 6, to the whole quadratic equation, and solve for $d$ .

We need a negative value of $d$ because as stated in the problem, the digits of the original number form a decreasing arithmetic sequence. But we cannot have $-6$ because it will leave the hundreds and the unit digit with a difference of $12$ , which is too large. This leaves us with $-4$ having the value of $d$ .

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Looking back to the equation for $a$ in terms of $d$ , we can now substitute the value of $d$ to find $a$ .

• $a = (\frac{d^2 + 2d}{2}$ )
• $a = (\frac{(-4)^2 + 2(-4)}{2}$ )
• $a = (\frac{16 - 8}{2}$ )
• $a = 4$ .

Finally, using the very first set of equations, we can now substitute the values of $a$ and $d$ to get each digit of the original number.

• Hundreds digit: $h = a - d$
• Hundreds digit: $h = 4 - (-4)$
• Hundreds digit: $h = 8$

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• Tens digit: $t = a$
• Tens digit: $t = 4$

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• Unit digit: $u = a + d$
• Unit digit: $u = 4 + (-4)$
• Unit digit: $u = 0$

And there we have it. The original number is $840$ . If we add it by 2, we will have $842$ . 8, 4, and 2 are geometrically sequenced.

$**Final Answer: 840**$

End Note: I know this is a very long solution. If you could provide a shortcut to this, it would be a great help.

Or if this solution is not worthy of your reading time, you may try to consider doing trial-and-error for the whole problem! :P