Difference between revisions of "2007 iTest Problems/Problem 13"
(Created page with "== Problem == What is the smallest positive integer <math>k</math> such that the number <math>{{2k}\choose k}</math> ends in two zeros? <math>\text{(A) } 3 \quad \text{(B) } 4 ...") |
Rockmanex3 (talk | contribs) (Solution to Problem 13) |
||
Line 17: | Line 17: | ||
\text{(M) } 2007\quad </math> | \text{(M) } 2007\quad </math> | ||
− | == Solution == | + | ==Solution== |
+ | |||
+ | When writing out <math>\binom{2k}{k}</math>, the numerator has the numbers from <math>k+1</math> to <math>2k</math> being multiplied, and the denominator has the numbers from <math>1</math> to <math>k</math> being multiplied. In order for <math>\binom{2k}{k}</math> to have two zeroes, the numerator must have two more factors of <math>2</math> and <math>5</math> than the denominator. | ||
+ | |||
+ | Going through the options from lowest to highest, the first value of <math>k</math> that satisfies the conditions is <math>13</math> because there are <math>4</math> factors of five and <math>12</math> factors of two in the numerator, while there are <math>2</math> factors of five and <math>10</math> factors of two in the denominator. The answer is <math>\boxed{\textbf{(K)}}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{iTest box|year=2007|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Introductory Number Theory Problems]] | ||
+ | [[Category:Introductory Combinatorics Problems]] |
Revision as of 03:27, 14 June 2018
Problem
What is the smallest positive integer such that the number
ends in two zeros?
Solution
When writing out , the numerator has the numbers from
to
being multiplied, and the denominator has the numbers from
to
being multiplied. In order for
to have two zeroes, the numerator must have two more factors of
and
than the denominator.
Going through the options from lowest to highest, the first value of that satisfies the conditions is
because there are
factors of five and
factors of two in the numerator, while there are
factors of five and
factors of two in the denominator. The answer is
.
See Also
2007 iTest (Problems, Answer Key) | ||
Preceded by: Problem 12 |
Followed by: Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4 |