while (a) {
b.push(a % 10);
a = Math.floor(a / 10);
if (b == 7) {
n = n + 1;
}
console.log("<br><br>number of 7's:" + n);
}
This is what I have come up with. The output is one of the numbers has seven; if not, then zero. I want the program to count the number of times seven appears in a number.
You can convert the number to a string, and then count how many times a character = 7:
let n = 7326577
let cnt = 0;
let strN = '' + n;
for(let c of strN)
if(c == '7')
cnt ++
console.log('Number of 7\'s in number: ' + cnt)Following you approach you need to store the last digit to a different variable and use that for checking if it is a 7
var a = 709728457;
var b = [];
var n = 0;
while (a) {
const lastDigit = a % 10;
b.push(lastDigit); // if you still need to store all digits
a = Math.floor(a / 10);
if (lastDigit == 7) {
n = n + 1;
}
}
console.log("number of 7's:" + n);var a = 7686774737
var no = String(a).split('').filter(e=>e==7).length;
console.log(no)