Divisibility, Remainders and Digits

You can test divisibility without dividing. For three or nine, add the digits. Seven thousand one hundred and twenty-eight has digit sum eighteen, so it is divisible by nine. For four, look at the last two digits: twenty-eight is a multiple of four. For eight, look at the last three digits: one hundred and twenty-eight is eight times sixteen. For eleven, alternate plus and minus along the digits: seven minus one plus two minus eight is zero, and zero is a multiple of eleven. To test for a number like seventy-two, test eight and nine, which share no common factor.

To find the units digit of a huge power, look for the cycle. Powers of seven end in seven, nine, three, one, and then seven again, because only the units digit decides the next units digit. So the cycle has length four. Two thousand and twenty-six divided by four is five hundred and six, remainder two. The remainder tells you the place in the cycle: the second place, which is nine. So seven to the power two thousand and twenty-six ends in nine.

Units digits are remainders when you divide by ten, and remainders by any number repeat in the same way. Sixty days after a Monday: sixty is eight whole weeks and four days, and whole weeks change nothing, so move four days on from Monday, to Friday. Now divide the powers of two by seven. Two, four and eight leave remainders two, four and one, and then the pattern repeats, so the cycle has length three. Forty divided by three is thirteen, remainder one, so two to the power forty leaves the first remainder in the cycle, which is two.

How many numbers from one to one hundred and fifty are divisible by three or five? There are fifty multiples of three and thirty multiples of five. But numbers like fifteen and thirty are in both lists, so they have been counted twice. They are the multiples of fifteen, and there are ten of them. Subtract them once: fifty plus thirty minus ten is seventy. The overlap is always the multiples of the lowest common multiple. For three and five that is fifteen, but for six and nine it is eighteen, not fifty-four.

Word problems hide the HCF and the LCM. Two bells ring every twelve and every eighteen minutes, and both ring at nine o'clock. They ring together again after a time that is a multiple of both, and the first such time is the lowest common multiple, thirty-six minutes, so at nine thirty-six. Now cut a ninety centimetre ribbon and a one hundred and fifty centimetre ribbon into equal pieces, as long as possible, with nothing wasted. The length must divide both, so it is the highest common factor, thirty centimetres, giving three plus five, which is eight pieces. Quiz and exercises now.