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**Section - 1****Ques 1. Is 4,005 divisible by 5?****Ans: **Yes: 4,005 ends in 5, so it is divisible by 5.**Ques 2. Does 51 have any factors besides 1 and itself?****Ans:** Yes: The digits of 51 add up to a multiple of 3 (5 + 1 = 6), so 3 is a factor of 51.**Ques 3. x = 20 ****The prime factors of x are:The factors of x are:Ans: **Prime factors: 2, 2, 5.

Factors: 1, 2, 4, 5, 10, 20

The prime factors of x are:

The factors of x are:

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The factors of x are:

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Factors: 1 ,2 ,4, 5 ,1 0 ,20 , 25, 50,100

The prime factors of x are:

The factors of x are:

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12 x 1 = 12

12 x 2 = 24

1 2 x 3 = 36

12 x 4 = 48

All other positive multiples of 12 are larger than 50.

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The factors of x are:

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Factors: 1, 2, 3, 6 ,7 ,14 ,21 ,42

The prime factors of x are:

The factors of x are:

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Even divisors can be built using one 2.

2; 2 x 3 = 6; 2 x 7 = 14; 2 x 3 x 7 = 42

Even divisors can also be built using both the twos that are prime factors of 84.

2 x 2 = 4; 2 x 2 x 3 = 12; 2 x 2 x 7 = 28; 2 x 2 x 3 x 7 = 84

The even divisors of 84 are 2, 4, 6, 12, 14, 28, 42, and 84.

Alternatively, you could make a factor pair table to see which factors are even:

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Factors: 1, 3, 13, 39

The prime factors of x are:

The factors of x are:

28 = 14 x 2 (that is, 28 contains everything that 14 contains, and 28 also has one additional factor of 2)

75 = 25 x 3 (that is, 75 contains everything that 25 contains, and 75 also has one additional factor of 3)

Therefore, the only additional prime factors in 28 x 75 are the 2 in 28 and the 3 in 75. Thus, the first product has two more prime factors than the second product.

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The factors of x are:

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Factors: 1,37

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Looking at the table, we can see that 45 and 15 are the two largest odd factors of 90.

Ques 24:Determine which of the following numbers are prime numbers. Remember, you only need to find one factor other than the number itself to prove that the number is not prime.

**2 3 5 6**

**7 9 10 15**

**17 21 27 29**

**31 33 258 303**

**655 786 1,023 1,325Ans:** Not prime: All of the even numbers other than 2 (6, 10, 258, 786), since they are divisible by 2.

All of the remaining multiples of 5 (15, 655, 1,325)

All of the remaining numbers whose digits add up to a multiple of 3, since they are divisible by 3, by definition: 9, 21 (digits add to 3), 27 (digits add to 9), 33 (digits add to 6), 303 (digits add to 6), and 1,023 (digits add to 6). All of these numbers are divisible by 3.

Ques 25: If x is divisible by 33, what other numbers is x divisible by?

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For x to be divisible by 12, we need to know that it contains all of the prime factors of 12. 12 = 2 x 2 x 3. Therefore 12 contains two 2s and a 3. x also contains two 2s and a 3, therefore xis divisible by 12.**Ques 28: If 7x is a multiple of 210, must x be a multiple of 12?Ans:** No: For x to be a multiple of 12, it would need to contain all of the prime factors of 12: 2, 2, and 3. If lx is a multiple of 210, it contains the prime factors 2, 3, 5, and 7. However, we want to know about x, not lx , so we need to divide out the 7. Therefore, x must contain the remaining primes: 2, 3, and 5. Comparing this to the prime factorization of 12, we see that x does have a 2 and a 3, but we don’t know whether it has two 2’s. Therefore, we can’t say that x must be a multiple of 12; it could be, but it doesn’t have to be.

Alternatively, we could start by dividing out the 7. If lx is divisible by 210, x is divisible by 30. We therefore know that x contains the prime factors 2, 3, and 5, and we can follow the remaining reasoning from above.

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Ques 30: If 40 is a factor of x, what other numbers are factors of x?

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For x to be divisible by 15, we need to know that it contains all of the prime factors of 15. 15 = 3 x 5. Therefore 15 contains a 3 and a 5. x also contains a 3 and a 5, therefore x is divisible by 15.

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We cant simply count all of the numbers that contain 2, because we might have some overlapping factors. For instance, 6 is a multiple of 2 and 3, so the fact that q is divisible by both 2 and 6 tells us only that we have at least one 2 (and at least one 3); we don’t necessarily have two factors of 2. Instead, we need to look for the largest number of 2’s we see in one factor. 12 contains two 2’s, so we know that q must be a multiple of 4, but we do not know whether q contains three 2’s. It might or it might not.

Alternately, we could run through our list of factors, adding to the list when new factors appear.

2: q must be divisible by 2.

6: The 3 is new. q must be divisible by 2 and 3.

9: The second 3 is new. q must be divisible by 2, 3, and 3.

12: The second 2 is new. q must be divisible by 2, 2, 3, and 3.

15: The 5 is new. q must be divisible by 2, 2, 3, 3, and 5.

30: Nothing new. q must be divisible by 2, 2, 3, 3, and 5.

Again, we see that we only have two 2s for certain. Therefore q must be a multiple of 180 (that is, 2 x 2 x 3 x 3 x 5), but it does not absolutely have to be a multiple of 8.

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What about the minimum? Can p and q have absolutely no factors in common? Try some numbers. If we choose p = 3 and q = 10, then the two numbers don’t have any prime factors in common, but notice that they are both divisible by 1. Any number is always divisible by 1. Therefore, our minimum possible number of factors is 1 (the number one itself) and our maximum is 2 (the two factors of prime number p>.

Ques 35: If 64 divides n, what other divisors does n have?

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For 12 to be a factor of n, « must contain all of the prime factors of 12. 12 = 2 x 2 x 3, so 12 contains two 2s and a 3. n also contains a 3 but only contains one 2 that we know of, so we don’t know whether 12 is a factor of n.**Ques 38: Positive integers a and b both have exactly four factors. If a is a one-digit number and b = a + 9, a =Ans:** We have a bit of a puzzle here. What kind of number has exactly four factors? Let’s start by looking at our most constrained variable— a. It is a positive one-digit number, so something between 1 and 9, inclusive, and it has four factors. We know that prime numbers have exactly two factors: themselves and one, so we only need to look at non-prime one-digit positive integers. That’s a small enough field that we can list them out:

1—just one factor!

4— 3 factors: 1, 2, and 4

6— 4 factors: 1, 2, 3, and 6

8— 4 factors: 1, 2, 4, and 8

9— 3 factors: 1, 3, and 9

So our two possibilities for a are 6 and 8. We now have to apply our two constraints for b. It is 9 greater than a, and it has exactly four factors. Here are our possibilities:

If a - 6, then b - 15. 15 has 4 factors: 1, 3, 5, and 15.

If a — 8, then b = 17. 17 is prime, so it has only has 2 factors: 1 and 17.

Only b - 15 works, so a must be 6.

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If n is the product of 2, 3, and 11, its factors are:

In this case, we can simply use the right-hand portion of our chart. We have four factors larger than 6: 11, 22, 33, and 66.

Notice that because the other given prime factors of n (2 and 3) multiply to get exactly 6, we can only get a number greater than 6 by multiplying by the third factor, the “two-digit prime number.” The right hand column represents that third factor multiplied by all of the other factors: 11 x 6, 11 x 3, 11 x 2, and 11 x 1. If we replace 11 with another two-digit prime, we will get the same result.

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For 30 to be a divisor of n, n has to contain all of the prime factors of 30. 30 = 2 x 3 x 5, so 30 contains 2, 3, and 5. n also contains 2, 3 and 5, so 30 is a divisor of n.**Ques 41: 4,2 1, and 55 are factors of n. Does 154 divide n?Ans:**

For 154 to divide n, n has to contain all the same prime factors as 154. 154 = 2 x 7 x 11, so 154 contains 2, 7, and 11. n also contains 2, 7 and 11, so 154 divides n.**Ques 42: If n is divisible by 196 and by 15, is 270 a factor of n?Ans:**

For 270 to be a factor of n, n must contain all the same prime factors as 270. 270 = 2 x 3 x 3 x 3 x 5, so 270 contains a 2, three 3 s, and a 5. n contains a 2 and a 5, but only one 3. Therefore, 270 is not definitely a factor of n.

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