Wiki User
∙ 14y agoCall your number 10x + y. x = y + 2 and 10x + y = 4 + 6(x + y)
Substitute y + 2 for x: 10(y + 2) + y = 4 + 6((y + 2) + y)
This simplifies to 10y + 20 + y = 4 + 6y + 12 + 6y, ie 20 - 16 = 12y - 11y so y = 4 and x = 6
Your number is 64, which is indeed 4 more than the sum of its digits.
Wiki User
∙ 14y ago27
415
27 2+7=9 9x3=27
543,210. If you can use them all multiple times, then it is 555,555.
Answer: 45 4 + 5 = 9 x 5 = 45
2
Eighteen
what is the greets possible 9 digit number that uses each of the digits 1-3 times
26.
45
804
27
21
27
27
-8
To count the number of times a digit occurs in an integer, start by initializing an array of ten counts of digits, such as int digits[10];Then, in a loop while the number is non zero, increment the element in the digits array that corresponds to the units digit, and then divide the number by ten, such as digits[number%10]++ and number/=10;int digits[10];int i;int number = some number;for (i=0; i