If 9/10 < a/b < 10/11, where a and b are positive integers, then find the smallest possible value of b.

bigmac Dec 10, 2019

#1**0 **

put the fractions in a common denominator (110)

99/110 < a/b < 100/110

does not help much, huh? There are no integers between 99 and 100 sooooo: Let's double everything (the next common denominator)

198/220 < a/b < 200/220 Now we might guess b to be least at 220 (and a to be 199)

ElectricPavlov Dec 10, 2019

#2**0 **

Doing some bashing we get the answer a/b as 19/21 which leaves the answer of b as 21.

goreisthebest Dec 12, 2019

#3**+1 **

This is how Gor.. came up with this answer.

9/10 < a/b < 10/11

so

18/20 < a/b < 20/22

There is a fair bet that 19/21 will be between the 2. (Maybe it is certainty, but I wouldn't want to say that without proving it)

Check by using decimals

9/10 = 0.9

10/11 = 0.909090....

19/21 = 0.90476

And it is between the 2 as predicted.

The smallest value of b is 21

Melody
Dec 12, 2019