I'm looking to calculate the square root of all numbers up to 9999. Are there any ways to instruct the program to skip numbers that do not have a perfect square root? Below is the current code I am using:
let i=1;
for (i===1;i>=1 && i <10000;i++){
let b = Math.sqrt(i);
console.log(`${i} = ${b}`);
}
To determine if the integer value matches the original value, you need to compare them.
let i=1;
for (i=1;i>=1 && i <10000;i++){
let b = Math.sqrt(i);
if (Math.trunc(b) == b)
console.log(`${i} = ${b}`);
}
Instead of using Math.trunc(b), consider using one of these alternatives:
Math.round(b)Math.floor(b)Math.ceil(b)parseInt(b, 10)Instead of testing every number up to 10000, you can calculate powers of two directly:
let count = 0, i = 0;
while (count < 10000) {
i++;
let b = i * i;
console.log(`${i} = ${b}`);
count = b;
}
Another way to approach this, as suggested in the comments, is to use a more elegant for-loop:
for (let i = 1; i*i < 10000; i++) {
console.log(`${i*i} = ${i}`);
}
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