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 January 4th, 2012, 06:09 PM #1 Newbie   Joined: Jan 2012 Posts: 28 Thanks: 0 Finding two digit number Can anyone help me with this problem? Five two-digit numbers are multiples of the products of their digits. Four of these numbers are 11, 12, 24, and 36. Find the fifth number.
 January 4th, 2012, 06:41 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond Re: Finding two digit number $10a\,+\,b\,=\,k\,\cdot\,a\,\cdot\,b\,\Rightarrow\, \frac{10}{b}\,+\,\frac{1}{a}\,=\,k$ If a = 1 and b = 5, k is an integer; answer: 15.
 January 4th, 2012, 07:44 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Finding two digit number Let x be the first digit (tens place) and y be the second digit (ones place). Since it is stated that there are only 5 such numbers, then we must exclude y = 0. We then require: $R_x=10x+y=nxy$ where $n\in\mathbb N$ and $0 $\frac{10x+y}{xy}=n$ $\frac{10}{y}=n-\frac{1}{x}$ $\frac{1\cdot2\cdot5}{y}=n-\frac{1}{x}$ We see that y may be 1, 2 or 5 when x = 1, so $R_1\in\{11,12,15\}$ We see that y may be 4 when x = 2, so $R_2\in{24}$ We see that y may be 6 when x = 3, so $R_3\in{36}$ Examination of the remaining values of x produce no other numbers satisfying the given conditions.

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