1
$\begingroup$

How many three digit numbers exist such that one of the digits is the geometric mean of the other two? A 12, B 18, C 24, D other

So, $N = 100a + 10b + c$ let $c =\sqrt{ab}$.

$ab$ must be a perfect square, so $ab = 1, 4, 9, 16, 25, 36, 49, 64, 81$

Case: $ab = 1 \implies a=1, b=1$.

Case: $ab = 4 \implies a = 2, b=2$

Case: $ab = 9 \implies a=3, b=3$.

Case: $ab = 16 \implies (a=8, b=2), (a=4, b=4)$

Case: $ab = 25 \implies a=5, b=5$

Case: $ab = 36 \implies (a=9, b=4), (a=6, b=6)$

Case: $ab = 49 \implies (a=7, b=7)$

Case: $ab = 64 \implies (a=8, b=8)$

Case: $ab=81 \implies (a=9, b=9)$.

$(a=8, b=2, c = 4)$ also means that we could replace this with, $ (a, b, c) = (2, 4, 8) = (4, 2, 8) = (4, 8, 2)$

$(a=9, b=4, c=6)$ also means $(a, b, c) = (9, 4, 6) = (4, 6, 9) = (9, 6, 4) = (4, 9, 6) = (6, 9, 4) = (6, 4, 9)$.

So to me, the answer is B $18$

$\endgroup$
  • 1
    $\begingroup$ You forgot the cases for $ab = 4$ where $a = 1, b = 4$ (or the other way around), and the same for the case $ab = 9$ $\endgroup$ – Arthur Jan 25 '16 at 12:14
  • 1
    $\begingroup$ Can't see the 6 permutations of 124 in there. $\endgroup$ – Frentos Jan 25 '16 at 12:21
  • 1
    $\begingroup$ $ab=16$ also implies the case of $a=2$ and $b=8$. $\endgroup$ – barak manos Jan 25 '16 at 13:43
  • $\begingroup$ BTW, the answer is $26$ (i.e., D - other). $\endgroup$ – barak manos Jan 25 '16 at 13:46
  • 2
    $\begingroup$ Already the title and the body of the post ask different questions... $\endgroup$ – Did Jan 25 '16 at 15:12
3
$\begingroup$

You asked two different questions. The question in the title appears to be:

"How many three digit numbers exist such that the third digit is the geometric mean of the other two?"

I will answer the question in the statement of the problem:

"How many three digit numbers exist such that one of the digits is the geometric mean of the other two?"

Let $x, y \geq 0$. The geometric mean of $x$ and $y$ is $\sqrt{xy}$.

A three digit positive integer is a number of the form $100a + 10b + c$, where $a \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ and $b, c \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

We consider cases:

  1. The geometric mean of the digits is $0 \implies 0 = \sqrt{a \cdot 0}$, where $a \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. There are nine possibilities: $100, 200, 300, 400, 500, 600, 700, 800, 900$.

  2. If $n \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $n = \sqrt{n \cdot n}$. There are nine such numbers. They are $111, 222, 333, 444, 555, 666, 777, 888, 999$.

  3. $2 = \sqrt{1 \cdot 4}$. Then the number is one of the $3! = 6$ permutations of the digits $1$, $2$, and $4$. They are $124, 142, 214, 241, 412, 421$.

  4. $3 = \sqrt{1 \cdot 9}$. Then the number is one of the $3! = 6$ permutations of the digits $1$, $3$, and $9$. They are $139, 193, 319, 391, 913, 931$.

  5. $4 = \sqrt{2 \cdot 8}$. The number is one of the $3! = 6$ permutations of the digits $2$, $4$, and $8$. They are $248, 284, 428, 482, 824, 842$.

  6. $6 = \sqrt{4 \cdot 9}$. The number is one of the $3! = 6$ permutations of the digits $4$, $6$ and $9$. They are $469, 496, 649, 694, 946, 964$.

Hence, there are $2 \cdot 9 + 4 \cdot 6 = 42$ three digit numbers such that one of the digits is the geometric mean of the other two.

$\endgroup$
-1
$\begingroup$

Third digit is $\sqrt{ab}$ implies square root should yield single digit so it can be $[0,9]$ thus now digits whose product is 0 so $100,..900$ , for 1 is only $1,1$ so number $111$ then for 2 so product is $4$ thus numbers are $222,142,412$ then $3$ so $333,193,913$ now $4$ so $284,824,444$ then $5$ so $555$ then $6$ so $496,946,666,$ for $7$ so $777$ then $8$ so $888$ and finally $9$ so $999$ thus total numbers are $9+1+3+3+3+1+3+1+1+1=26$ so its $D$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.