For how many integer values of \(n\) between 1 and 349 inclusive does the decimal representation of \(n\over350\) terminate?

Jmaster10 Jan 22, 2021

#1**-1 **

I wrote a computer program

i = 0

for (n = 1 to 349)

if terminate(n/350) == 1

i = i + 1

write(output,i)

output: i = 40

There are 40 numbers n that work.

bigmac Jan 22, 2021

#2**+2 **

Here's an answer without computer programming.

first, find the prime factorization of 350. (2*5^2*7)

then we need to find all multiples of 7 that are between 1 and 350.

Solution 1: Count.

7 14 21 28 35 42 49 56 63 70 77 84 91 98 105 112 119 126 133 140 147 154 161 168 175 182 189 196 203 210 217 224 231 238 245 252 259 266 273 280 287 294 301 308 315 322 329 336 343 350

There. Now we count the numbers. After counting, you see that there are **50**.

Solution 2: Find what is the first multiple of 7/350.

The first multiple of 7 between 1 and 350 is 7. so 7/350=0.02.

Then, drop the two zeros and find out what is 100% (1 whole, ignoring the percent) /2.

Then, we get **50**.

Also, @Guest is wrong.

KitSoundwave Jan 22, 2021