Euclid 2019/Problem 3

Problem

(a) Michelle calculates the average of the following numbers: $5, 10, 15, 16, 24, 28, 33, 37$ . Daphne removes one number and calculates the average of the remaining numbers. The average that Daphne calculates is one less than the average that Michelle calculates. Which number does Daphne remove?

(b) If $16^{\frac{15}{x}}=32^{\frac43}$ , what is the value of $x$ ?

(c) Suppose that $\frac{2^{2022}+2^a}{2^{2019}}=72$ . Determine the value of $a$ .

Solution

(a) The average of the original set is $\frac{5+10+15+16+24+28+33+37}{8}=21$ , so the new average must be $21-1=20$ . Because the original sum of the numbers was $168$ and the new sum of the numbers is $20\times 7=140$ , so the number taken off must be $168-140=\boxed{28}$ .

(b) After writing both sides of the equation in powers of $2$ , we have $2^{4^{\frac{15}{x}}}=2^{5^{\frac43}}$ , so $2^{\frac{60}{x}}=2^{\frac{20}{3}}$ , so $\frac{60}{x}=\frac{20}{3}$ , so $180=20x$ , so $x=\boxed{9}$ .

(c) Let's write the left side as $\frac{2^{2022}}{2^{2019}}+\frac{2^a}{2^{2019}}=2^3+2^{a-2019}=8+2^{a-2019}$ . So, after subtracting $8$ from both sides, we will get $2^{a-2019}=64$ , so $2^{a-2019}=2^6$ , so $a-2019=6$ , so $a=\boxed{2025}$ .

~Yuhao2012

Page

Toolbox

Search

Euclid 2019/Problem 3

Problem

Solution