Difference between revisions of "1985 AJHSME Problem 9"
Coolmath34 (talk | contribs) (Created page with "== Problem == The product of the 9 factors <math>\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =</math> <math>\text{(A)}\ \...") |
(→Solution 2 (Brute Force)) |
||
| (6 intermediate revisions by 2 users not shown) | |||
| Line 3: | Line 3: | ||
<math>\text{(A)}\ \frac {1}{10} \qquad \text{(B)}\ \frac {1}{9} \qquad \text{(C)}\ \frac {1}{2} \qquad \text{(D)}\ \frac {10}{11} \qquad \text{(E)}\ \frac {11}{2}</math> | <math>\text{(A)}\ \frac {1}{10} \qquad \text{(B)}\ \frac {1}{9} \qquad \text{(C)}\ \frac {1}{2} \qquad \text{(D)}\ \frac {10}{11} \qquad \text{(E)}\ \frac {11}{2}</math> | ||
| + | |||
| + | == In-depth Solution by BoundlessBrain!== | ||
| + | https://youtu.be/yBrLXnjasgg | ||
== Solution == | == Solution == | ||
| Line 8: | Line 11: | ||
<cmath>\frac{1}{2} \cdot \frac{2}{3} \dots \frac{9}{10}.</cmath> | <cmath>\frac{1}{2} \cdot \frac{2}{3} \dots \frac{9}{10}.</cmath> | ||
Numerators and denominators cancel to yield the answer: <math>\boxed{\text{(A)} \frac{1}{10}}.</math> | Numerators and denominators cancel to yield the answer: <math>\boxed{\text{(A)} \frac{1}{10}}.</math> | ||
| + | |||
| + | == Solution 2 (Brute Force) == | ||
| + | |||
| + | Multiply the numerators and denominators | ||
| + | |||
| + | The expression is <math>\frac{362880}{3628800}</math> which is <math>\frac{1}{10}</math> or <math>\boxed{\textbf{(A)}\frac{1}{10}}</math> | ||
| + | |||
| + | |||
| + | <math>\textbf{Note: Do not use unless no other method found}</math> | ||
Latest revision as of 08:29, 6 October 2024
Problem
The product of the 9 factors
In-depth Solution by BoundlessBrain!
Solution
This product simplifies to:
Numerators and denominators cancel to yield the answer:
Solution 2 (Brute Force)
Multiply the numerators and denominators
The expression is
which is
or