If the sum of the digits in a two-digit number is 15, and the number is 6 more than 15 times the units digit, the number is 96. Let A and B be the digits. (A is the tens digit and B is the units digit) A + B = 15, therefore A = 15 - B 10A + B = 15B + 6
Substituting for A, and solving for B...
10(15 - B) + B = 15B + 6
150 - 10B + B = 15B + 6
150 - 9B = 15B + 6
144 - 9B = 15B
144 = 24B
B = 6 Back substitute B into first equation and solve for A...
A + 6 = 15
A = 9 Therefore, the digits are 9 and 6, so the number is 96.
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
The two digit numbers in which the tens digit is five greater than the units digit are:5061728394The number we are looking for is ten times the sum of its digits, and so must be a multiple of 10. The only one of the above numbers which fits this is 50, as 5 is five greater than 0, and 50 = 10 x (5+0).
Call the digits 'A' and 'B', so that the number is written: AB.10A + B = 4 (A+B)10A + B = 4A + 4B6A = 3B2A = B ---> any 2-digit number where the units digit is twice the tens digit12 or 24 or 36 or 48
The last two digits (the tens and units) are divisible by 4.This is equivalent to the following two conditions:If the tens digit is even, the units digit must be 0, 4 or 8If the tens digit is odd, the units digit must be 2 or 6For divisibility by 9, calculate the digital root: this is the sum of all the digits in the number. Repeat with the digits of this number - and keep repeating until you are down to a single digit. It that is 9, then the number is divisible by 9 and if not, it is not.
As the digits are moved left, the digit in the tenths column goes into the units column, the digit in the hundredths column goes into the tenths column, etc; each digit is ten times its previous value, thus moving the digits to the left multiplies the number by 10. Similarly moving the digits to the right: the digit in the units column goes into the tenths column, the digit in the tenths column goes into the hundredths column, etc; each digit is a tenth of its previous value, thus moving the digits to the right divides the number by 10.
-4
Any two digit number in which: (a) the units digit is not 0, and (b) the two digits are different will form a new 2-digit number when the digits are interchanged.
The units digit of a two digit number exceeds twice the tens digit by 1. Find the number if the sum of its digits is 10.
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
The two digit numbers in which the tens digit is five greater than the units digit are:5061728394The number we are looking for is ten times the sum of its digits, and so must be a multiple of 10. The only one of the above numbers which fits this is 50, as 5 is five greater than 0, and 50 = 10 x (5+0).
What is the units digit of the least whole number greater than 1000 whose digits are all different?
17
Call the digits 'A' and 'B', so that the number is written: AB.10A + B = 4 (A+B)10A + B = 4A + 4B6A = 3B2A = B ---> any 2-digit number where the units digit is twice the tens digit12 or 24 or 36 or 48
I am a three digit number. all of my digit is multiples of 2. my hundreds digit is the lowest even number. the sum of my digits is 16. the units digit is the same as the difference between my hundreds and tens digits.
The last two digits (the tens and units) are divisible by 4.This is equivalent to the following two conditions:If the tens digit is even, the units digit must be 0, 4 or 8If the tens digit is odd, the units digit must be 2 or 6For divisibility by 9, calculate the digital root: this is the sum of all the digits in the number. Repeat with the digits of this number - and keep repeating until you are down to a single digit. It that is 9, then the number is divisible by 9 and if not, it is not.
As the digits are moved left, the digit in the tenths column goes into the units column, the digit in the hundredths column goes into the tenths column, etc; each digit is ten times its previous value, thus moving the digits to the left multiplies the number by 10. Similarly moving the digits to the right: the digit in the units column goes into the tenths column, the digit in the tenths column goes into the hundredths column, etc; each digit is a tenth of its previous value, thus moving the digits to the right divides the number by 10.
If the last two digits are divisible by 4 then the number is divisible by 4. Thus, if the tens digit is even and the units digit is 0 or 4 or 8 OR if the tens digit is odd and the units digit is 2 or 6 then the number is divisible by 4.