Difference between revisions of "1985 AJHSME Problem 9"

Revision as of 07:28, 6 October 2024

Problem

The product of the 9 factors $\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =$

$\text{(A)}\ \frac {1}{10} \qquad \text{(B)}\ \frac {1}{9} \qquad \text{(C)}\ \frac {1}{2} \qquad \text{(D)}\ \frac {10}{11} \qquad \text{(E)}\ \frac {11}{2}$

In-depth Solution by BoundlessBrain!

https://youtu.be/yBrLXnjasgg

Solution

This product simplifies to: $\frac{1}{2} \cdot \frac{2}{3} \dots \frac{9}{10}.$ Numerators and denominators cancel to yield the answer: $\boxed{\text{(A)} \frac{1}{10}}.$

Solution 2 (Brute Force)

Multiply the numerators and denominators The expression is $\frac{362880}{3628800}$ which is $\frac{1}{10}$ or $\boxed{\textbf{(A)}\frac{1}{10}}$

Revision as of 07:27, 6 October 2024 (view source) Lovelearning999 (talk \| contribs) ← Older edit		Revision as of 07:28, 6 October 2024 (view source) Lovelearning999 (talk \| contribs) (→‎Solution 2 (Brute Force)) Newer edit →
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	Multiply the numerators and denominators		Multiply the numerators and denominators
−	The expression is <math>\frac{362880}{3628800}</math> which is <math>\frac{1}{10}</math>	+	The expression is <math>\frac{362880}{3628800}</math> which is <math>\frac{1}{10}</math> or <math>\boxed{\textbf{(A)}\frac{1}{10}}</math>

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