Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$

Question

This question appeared in one of the national exams (MCQs) in Saudi Arabia.

In this exam;

• Using calculators is not allowed,
• The student have $72$ seconds on average to answer one question.

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PROBLEM:

Compare $a=(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $b=(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$.

CHOICES:

A) $a>b$

B) $a<b$

C) $a=b$

D) Given information is not enough

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Using algebra to evaluate each expression is easy, and the correct choice is $A$, but that will take a long time.

Any suggestion to solve this problem in a short time? THANKS.

Answer 1

I do not think it would take a long time to simplify these expression by hand. Rewrite it as $$\frac {5}{2}\cdot\frac {10}{3}\cdot\frac {17}{4}\cdot\frac {26}{5}\text{ vs }\frac {11}{5}\cdot \frac {13}{4}\cdot\frac {13}{3}\cdot \frac {11}{2}$$ Denominators go away, and the factor $13$ in the numerator as well: $$1700\text{ vs }13\cdot 121,$$ and the latter seems to be easy to estimate.

Answer 2

If $a<b$ then $$(a+x)(b-x)$$ is increasing in x for $0\leq x \leq \frac{b-a}{2}$.

Using this $(2+1/2)*(5+1/5)$ is larger than $(2+1/5)*(5+1/2)$ and $(3+1/3)*(4+1/4)$ is larger than $(3+1/4)*(4+1/3)$.

Intuitively, the square maximises the area over all rectangles with the same circumference. To maximise a product where the factors have a fixed sum, we must try to get the factors as close as possible.

Answer 3

The sum of all four factors is the same in both cases. To maximize the product, we want the factors to be as close together as possible.

The factors $(2 + \frac12)(5 + \frac15)$ are closer to their average than $(2 + \frac15)(5 + \frac12)$, so $(2+\frac12)(5+\frac15) > (2+\frac15)(5+\frac12)$.

Similarly, $(3 + \frac13)(4+\frac14) > (3 + \frac14)(4 + \frac13)$.

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We could also compare each of these pairs by multiplying them out. But we don't have to multiply out everything. When we expand $(2 + \frac12)(5 + \frac15)$ and $(2 + \frac15)(5 + \frac12)$, the terms $2\cdot 5$ and $\frac12 \cdot \frac15$ will be common between them. However, $2 \cdot \frac15 + \frac12 \cdot 5 > 2 \cdot \frac12 + \frac15 \cdot 5$, which is not hard to see: just $\frac12 \cdot 5$ is $2.5$ on the left, and the right is $2$.

The same thing happens with the other pair: $3 \cdot \frac14 + \frac13 \cdot 4 > 3 \cdot \frac13 + \frac14 \cdot 4$.

Answer 4

We have that

$$a=\left(2+\frac{1}{2}\right)\left(3+\frac{1}{3}\right)\left(4+\frac{1}{4}\right)\left(5+\frac{1}{5}\right)=$$

$$=\frac12\left(4+1\right)\frac13\left(9+1\right)\frac14\left(16+1\right)\frac15\left(25+1\right)=$$

$$=\frac{5\cdot 10\cdot17\cdot26}{120}$$

and similarly

$$b=\left(2+\frac{1}{5}\right)\left(3+\frac{1}{4}\right)\left(4+\frac{1}{3}\right)\left(5+\frac{1}{2}\right)=\frac{11\cdot 13\cdot13\cdot11}{120}$$

with

$$5\cdot 10\cdot17\cdot26 > 11\cdot 13\cdot13\cdot11$$ $$5\cdot 10\cdot 17\cdot 2 > 11\cdot 13\cdot11$$

$$100 \cdot 17 > 121 \cdot 13$$

Answer 5

My intuition is to notice that

$$\frac{2+\frac{1}{2}}{2+\frac{1}{5}}\ ?\ \frac{5+\frac{1}{2}}{5+\frac{1}{5}}$$

(I have used $?$ since I do not know how these expressions relate)

$$\frac{5}{2}\frac{5}{11}\ ?\ \frac{11}{2}\frac{5}{25}$$

$$\frac{25}{22}\ ?\ \frac{55}{50}$$

$$\frac{50}{44}\ ?\ \frac{55}{50}$$

Since $50-44=6>5=55-50$, we know that $?$ is actually $>$. Then

$$\left(\frac{2+\frac{1}{2}}{2+\frac{1}{5}}\right)(3+\frac{1}{3})(4+\frac{1}{4})>(2+\frac{1}{5})(3+\frac{1}{4})\left(\frac{5+\frac{1}{2}}{5+\frac{1}{5}}\right)$$

and the expression simplifies, giving answer $A$.