Multiples and divisors

Division Theorem: For every pair of integers and , there exist unique integers and such that , where , is the quotient and is the remainder. The integer is divisible by integer if the remainder is If is divisible by ( divides is a multiple of and is a factor (divisor) of

Divisibility rules:

·         A number is divisible by 2 (even number) if the last digit is divisible by 2.

·         A number is divisible by 3 if the sum of its digits is divisible by 3.

·         A number is divisible by 4 if the last two digits are divisible by 4.

·         A number is divisible by 5 if the units digit is 0 or 5.

·         A number is divisible by 6 if it is an even number divisible by 3.

·         A number is divisible by 7 if we double the last digit and subtract it from the rest of the numbers to yield a number divisible by 7.

·         A number is divisible by 8 if the last three digits are divisible by 8.

·         A number is divisible by 9 if the sum of its digits is divisible by 9.

·         A number is divisible by if its last digits are divisible by A number is divisible by if its last digits are divisible by A number is divisible by if its last digits are divisible by

A prime number is a natural number greater than 1 whose only factors are 1 and itself. A composite number is a natural number that has at least one factor besides 1 and itself. is the only prime number that is even. The following are the prime numbers below 100:

If a natural number greater than 1 has no prime divisors less or equal to is a prime number. If is a composite number greater than 1, the largest prime factor of is less or equal to .

We can write an integer as the product of the prime numbers it is divisible by. Prime factorization finds which prime numbers multiply together to make the original integer n: , where the factors are all prime numbers.

For example, the prime factorization of is .

Rules of odd and even number operations:

odd + odd = even, even + even = even, odd + even = odd, even + odd = odd

odd – odd = even, even – even = even, odd – even = odd, even – odd = odd

odd × odd = odd, even × even = even, odd × even = even, even × odd = even

Prove that a number is divisible by 9 if the sum of its digits is divisible by 9.

Solution: Let’s express the original integer as . We want to prove that if ( is an integer), then a is divisible by 9. For any , let We have:

This is always divisible by 9, since and are all divisible by 9. If is divisible by 9, we have and must be divisible by 9 as well.

Similarly, we can show that is the necessary and sufficient condition for to be divisible by 11. We start with the units digit and add every other number and then subtract the sum of the remaining numbers. If the result is divisible by 11, is divisible by 11.

General Rule: Whenever we need to use the sum (or alternate sum or difference) of the digits in an integer, express the integer in expanded form as

is an integer between 1 and 100,000. When we divide by 1,000, is the quotient and is the remainder. If is divisible by 111, how many possible values can take?

Solution: By definition, we have We need to identify the a’s that make divisible by 111. Note that Since is divisible by 111, is divisible by 111. That means if is divisible by 111, then is divisible by 111. Thus, we only need to identify the number of multiples of 111 between 1 and 100000. There are 900 multiples of 111 between 1 and 100000. Therefore, the number of possible values that can take is 900.

Part A. is a 4-digit integer that is divisible by 9. How many possible values can take if the last two digits of are 25?

Part B. is a 4-digit integer that is divisible by 9. How many possible values can take if the hundreds digit is 2 and the tens digit is 5?

Part C. is a 5-digit integer that is divisible by 3. How many possible values can take if the last two digits of are 25?

Solution: For part A, let’s express as where is the thousands digit and is the hundreds digit. For to be divisible by 9, we need to have divisible by 9, which means that when we divide by 9, the remainder needs to be 2. Since is between 1 and 9 inclusive and is between 0 and 9 inclusive, the possible values of are 2 and 11. If is 2, we have 2 choices for (20 or 11). If is 11, we have 8 choices for (29, 38, …, 83, 92). Therefore, there are 10 possible choices altogether. Another approach is to treat as a two-digit number between 10 and 99 inclusive. That the remainder of divided by 9 is 2 is equivalent to that the remainder of divided by 9 is 2. The reason is that Therefore, we are essentially counting the number of two-digit numbers that have a remainder of 2 when divided by 9. There are 10 of them: 11, 20, …, 83, 92.

General Rule: When we check for divisibility by 3 or 9, we can not only detach any digits from the number but also combine any digits.

For example, a 4-digit number which is divisible by 9 is equivalent to all of the following:

is divisible by 9 (Standard divisibility test)

is divisible by 9 (Part A of the problem)

is divisible by 9 (Yes, we can switch the position of the digits.)

is divisible by 9 (Now we have part B of the problem.)

Therefore, part B is the same as part A. We are essentially counting the number of two-digit numbers that have a remainder of 2 when divided by 9. The answer is again 10.

Part C is just another application of the general rule. Let’s express as Since the remainder of 25 divided by 3 is 1, we need to count the number of 3-digit numbers () between 100 and 999 inclusive that have a remainder of 2 when divided by 3. 300 numbers[2] satisfy the requirement: 101, 104, …, 995, 998. Therefore, the answer is 300.

and are two prime numbers. If and are also prime numbers, what are the values of and ?

Solution: Since the sum of two odd numbers is an even number and the difference of two odd numbers is also an even number, a and cannot be both odd numbers. The only prime number that is even is 2. Therefore, Among and one of them must be a multiple of 3.[3] If a number is a multiple of 3 and is also a prime number, that number is 3 itself. Therefore, we should have and We can verify that is also a prime number.

General Rule: If the sum or the difference of two prime numbers is odd, one of the prime numbers must be If the product of two or more prime numbers is even, at least one of the prime numbers is

How many factors does 720 have?

Solution: Since the prime factorization of 720 is , each factor of 720 can be expressed as . There are 5 choices for a (0, 1, 2, 3, 4), 3 choices for b (0, 1, 2), and 2 choices for c (0, 1). The choices of a, b, and c are independent. For example, if and , we get one factor, if and , we get another factor, The total number of choices is Therefore, 720 has 30 factors.[4]

General Rule: If the prime factorization of a number is , where are distinct prime numbers and are positive numbers, then the total number of factors that has is

How many positive integers less than 1000 have exactly 9 positive integer factors?

Solution: Using the general rule from the last problem, we know that if , it has factors. If an integer has 9 factors, we have There are two possible choices: and For the first choice, where is a prime number. Among all prime numbers, only is less than 1000. For the second choice, where and are different prime numbers. In other words, is the perfect square of a number that is the product of two distinct prime numbers. Since and the maximum value of the product of the two different prime numbers is 31. Let’s check what the possible combinations are:

If can be (5 possible values)

If can be (2 possible values)

No other prime number pairs satisfy the requirement. Therefore, the number of positive integers less than 1000 that have exactly 9 positive integer factors is

General Rule: A positive integer is a perfect square if its total number of factors is an odd number. This conclusion is straightforward as must all be even numbers in the prime factorization if is a perfect square. If the total number of factors is a prime number the only choice is . Therefore, is the power of a prime number.

If the largest prime factor of is and the largest prime factor of is what are the possible values of ?

Solution: Let and Both and are prime numbers and we also have Thus, and must be pairs of divisors of 72. If the sum of two integers is odd, their difference must be odd; If the sum of two integers is even, their difference must be even. Since the product of and is even, both must be even. Since the following are three possible pairs:

. This pair works and

This pair works and

. This pair does not work as is not a prime number.

Therefore, possible values of are 49 and 289.

General Rule: If the sum of two integers is odd, their difference must be odd. If the sum of two integers is even, their difference must be even.

Part A. What is the units digit of

Part B. What is the units digit of

Solution: The units digit of repeats the pattern of as increases . All , where is an integer, has the same units digit as . Thus, the units digit of is . The units digit of repeats the pattern of as increases. All has the same units digit; has the same units digit as . The units digit of is always 5. The units digit of of is the same as the units digit of Therefore, the units digit is

For part B, we need to be more careful with the sign of Since if is positive, the equation borrows from the tens digit to make the units digit 4. Is positive? Let’s quickly estimate. Since is much smaller than the focus is on versus . Clearly Therefore, is positive and the units digit is

General Rule: The units digit (ones digit) of the sum, difference, or product of integers only depend on the units digits of all the integers used in the calculation. The tens digit of the sum, difference, or product of integers solely depend on the tens digit and the units digits of all the integers used in the calculation. In general, the last n digits of the sum, difference, or product of integers only depend on the last n digits of the integers used in the calculation.

All the integers between 1 and 99 inclusive, excluding those that are multiples of 5, are multiplied together. What is the units digit of the product?

Solution: For single-digit numbers, we can track the units digits of Remember that we are not doing the full multiplication; only tracking the units digit of each multiplication as the units digits of the products solely depends on the units digits of the multipliers:

Similarly, we only need to track the unit digits when the numbers become two digits: the product of 8 numbers from 11 to 19 (21 to 29, 31 to 39, …, 91 to 99) yields 6 as the units digit as well. Since after each group of multipliers are included, the units digit remains as 6. Therefore, the units digit of the final product is 6.

What is the tens digit of ? What is the tens digit of ?

Solution: If the solution does not come naturally, let’s begin with calculating with increasing Since the tens digit of the product of two integers only depends on their tens digit and the units digit, we only need to track the last two digits of

Now we can see a clear pattern. The last two digits repeat the pattern of every four powers. Therefore, the tens digits of is the same as the tens digit of which is Let’s try the same approach for

The pattern for is different. We do not have repeated. Starting from the last two digits repeat the pattern of every five powers. Therefore, the tens digit of is the same as the tens digit of which is

is a 2-digit number. If we switch the tens digit and units digit of the sum of the new number and is a perfect square (e.g., . How many possible values of are there?

Solution: Let’s express as where is an integer between 1 and 9 and is an integer between 0 and 9. Then we have and If is a perfect square, must be a multiple of . Since the maximum value of is we have The possible values of are Therefore, the number of possible values of is 8.

General Rule: Whenever we need to switch the digits of a number express as

If you reverse the digits in a 5-digit number (e.g., to ), which of the following may be the difference between the reversed number and the original number?

(A) 34567 (B) 21092 (C) 14949 (D) 54319 (E) 42102

Solution: We need to identify the characteristic of the difference between a reversed number and its original number to eliminate four of the five choices. Let’s express the original number as The reverse number is The difference must be a multiplier of 9 and a multiplier of 11.

Since checking divisibility by 9 is easier, let’s begin with summing up the digits in each choice and eliminate the ones that do not add up to a multiplier of 9. This step eliminates A, B, and D. For a 5-digit number to be divisible by 11, the difference between the sum of the first, the third, and the fifth digit and the sum of the second and the fourth digit must be a multiplier of 11. E does not meet the requirement while C does Therefore, the answer is C.