Difference between revisions of "1999 CEMC Pascal Problems/Problem 14"

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-==Problem==
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-For how many different values of <math>k</math> is the <math>4</math>-digit number <math>7k52</math> divisible by <math>12</math>?
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-<math> \text{ (A) }\ 0 \qquad\text{ (B) }\ 1 \qquad\text{ (C) }\ 2 \qquad\text{ (D) }\ 3 \qquad\text{ (E) }\ 4</math>
-==Solution==
-We can use the fact that a number that is divisible by <math>12</math> must be divisible by both <math>3</math> and <math>4</math>, since <math>12 = 3 \times 4</math>.
-The divisibility rule for <math>4</math> states that that the last two digits must be divisible by <math>4</math>. Since <math>52 / 4 = 13</math>, which is an integer, we know that the number must be divisible by <math>4</math>, no matter what the value of <math>k</math> is.
-The divisibility rule of <math>3</math> states that the sum of the digits of the number must be divisible by <math>3</math>.
-Summing up the digits of the number, we get
-<math>7 + k + 5 + 2 = k + 14</math>
-This is divisible by <math>3</math> when <math>k = 1</math>, so it is divisible by <math>3</math> when <math>k = 4</math> and <math>k = 7</math>. <math>k</math> is a digit, so it must be an integer between <math>1</math> and <math>9</math> (inclusive)
-This gives <math>\boxed {\textbf {(D) } 3}</math> possible values of <math>k</math>.
-~anabel.disher