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Would you have made it into Harvard in 1869?

Posted on May 6, 2011 by Guillermo Bautista | 1 Comment

If you think your tough in math and can pass the 1869 Harvard math exam, don’t rejoice so much because there are other subjects as well.

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Other subjects included are History and Geography, Latin Grammar, English to Latin Translation, Greek Grammar with accents, and Greek Composition .

Surprisingly, according to the original article, 185 out of 210 examinees passed the exam.

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Posted in Math Lite, Problems Sets

Tagged harvard 1869 entrance exam, harvard 1869 sample exam, harvard sample exam, harvard sample exam in greek, harvard sample exam in latin, ivy league sample exam

Problem Set 2

Posted on October 13, 2009 by Guillermo Bautista | Leave a comment

PROBLEMS

1.) Find a linear function $f(x)$ such that $f(1) = 42$ and $f(2) = 47$ .

2.) Solve for $x$ : $4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 170.$

3.) Prove that the product of $3$ consecutive numbers is always divisible by $6$ .

4.) Prove that if $p$ is prime, $a$ and $b$ are integers, and $a \equiv b\mod p$ , then $a^p \equiv b^p \mod p$ .

SOLUTIONS AND PROOFS

Post Date: October 20, 2009

1. Solution: This is just the same as saying, find the equation of the line passing through $(1,42)$ and $(2,1337)$ . So, by point slope formula, we have, $y - y_1 = \displaystyle\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\Rightarrow y - 42 = \displaystyle\frac{1337 - 42}{2 - 1}(x - 1). \Rightarrow y = f(x) = -18x + 92.$

2.) Solution: $4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 4^x(1 + 4 +4^2 +4^3) = 170 \Rightarrow 4^x(85) = 170\Rightarrow 4^x = 2 \Rightarrow 2^{2x}=2^1 \Rightarrow x = \displaystyle\frac{1}{2}.$

3.) Proof: A number is divisible by $6$ if it is divisible by $2$ and $3$ . A product of $3$ consecutive numbers is divisible by $2$ because at least one of them is even, so it remains to show it is divisible by $3$ .

If a number is divided by $3$ , its possible remainders are $0, 1,$ and $2$ . Assume $n, n +1$ and $n+2$ be the three consecutive numbers, and $r$ be the remainder if $n$ is divided by $3$ .

Case 1: If $r=0$ , we are done.

Case 2: If $r = 1$ , then $n + 2 \Rightarrow r=0$

Case 3: If $r = 2$ , then $n + 1 \Rightarrow r = 0$ .

Since the product of the three consecutive numbers is even, and for each case of $r$ , one of the consecutive numbers is divisible by $3$ , the product of three consecutive numbers is divisible by $6. \blacksquare$

4.) Proof: From definition, $a^p \equiv b^p \mod p \Leftrightarrow b = a + kp$ for some $k \in \mathbb{Z}.$

Raising both sides of the equation to $p$ , we have $b^p = (a + kp)^p.$ By the binomial theorem, $b^p = (a + kp)^p = a^p + \displaystyle {p \choose 1}a^{p-1}kp + \displaystyle{p \choose 2}a^{p-2}k^2p^2 + \ldots + k^pp^p$ .