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Number and Operations Session 5: Solutions
 
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A B C 
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Solutions for Session5, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6 | B7 | B8 | B9 | B10 | B11


Problem B1

a. 

All numbers ending in 2, 4, 6, 8, or 0 are multiples of 2.

b. 

All numbers ending in 5 or 0 are multiples of 5.

c. 

All numbers ending in 0 are multiples of 10.

d. 

The test for divisibility by any of these numbers tests whether the units digit of the given number is divisible by 2, 5, or 10. You only need to look at the units digit, since any tens digit is automatically a multiple of 2, 5, and 10. For this same reason, all digits higher than the units digit can be ignored. We'll explore why this works later in this part of the session.

<< back to Problem B1


 

Problem B2

a. 

The following number are divisible by 9: 9, 18, 27, 36, 153, 2,466.

b. 

The following numbers are divisible by three: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 153, 156, 159, 2,463, 2,466, and 2,469.

c. 

One test for divisibility by 9 or 3 is to add the digits of the number being used and then test that for divisibility by 9 or 3. So, for example, 2 + 4 + 6 + 6 = 18, so 2,466 is divisible by 9. We'll explore why later in this part.

<< back to Problem B2


 

Problem B3

Since a number is divisible by 6 only when it is divisible by both 2 AND 3, we can combine the rules for 2 and 3: A number is divisible by 6 when its units digit is divisible by 2 (i.e., it is 2, 4, 6, 8, or 0) and when the sum of its digits is divisible by 3.

<< back to Problem B3


 

Problem B4

Using 2 and 3 to check the divisibility by 6 is possible because 2 and 3 are relatively prime; that is, the only factor they have in common is 1.

Similarly, the divisibility test for 15 is constructed from the divisibility tests for 3 and 5 (they are also relatively prime): A number is divisible by 15 when its units digit is divisible by 5 (i.e., it is 5 or 0) and when the sum of its digits is divisible by 3.

<< back to Problem B4


 

Problem B5

a. 

To test for divisibility by 4, we need to check the number that comprise the units and tens digit of the number. So to test whether 32,464 is divisible by 4, we test 64 to see if it's divisible by 4. It is, so 32,464 is divisible by 4.

To test for divisibility by 8, you need to check the numbers that comprise the last three digits of the number. So, to test whether 32,464 is divisible by 8, we test 464 to see if it's divisible by 8. It is, so 32,464 is divisible by 8.

b. 

Likewise, to test 82,426, we see if 26 is divisible by 4; it isn't, so 82,426 is not divisible by 4.

Since 82,426 is not divisible by 4, it can't possibly be divisible by 8 either, so we don't need to apply the test.

<< back to Problem B5


 

Problem B6

In the number 3,456, the digits for even powers of 10 are 4 and 6. The digits for odd powers of 10 are 3 and 5. The sum for the even powers is 10, and the sum for the odd powers is 8. The difference between the sums is 2. Since 2 is not divisible by 11, 3,456 is not a multiple of 11.

<< back to Problem B6


 

Problem B7

a. 

For this number to be divisible by 9, the sum of its digits must be a multiple of 9. If we call the missing digit x, then 1 + 2 + 4 + 7 + 3 + x = (17 + x) must be a multiple of 9. The only possible value of x that satisfies this is 1, so the number is 124,731, a multiple of 9.

b. 

For this number to be divisible by 33, it must satisfy the tests for 3 and 11. Let's look at 11 first; we need the sum of the odd and even powers of 10:

Even powers: 6 + 1 + x = 7 + x
Odd powers: 3 + 4 + 2 = 9

The difference between these must be a multiple of 11, and therefore the only possible value for x is 2. The number 364,122 is a multiple of 11.

To check whether it is a multiple of 3, add its digits: 3 + 6 + 4 + 1 + 2 + 2 = 18, a multiple of 3. Therefore 364,122 is a multiple of 33, since it is a multiple of both 3 and 11.

<< back to Problem B7


 

Problem B8

No! If a number is a multiple of 6, then we already know it is a multiple of 2. This makes the test for 2 irrelevant and unhelpful. In general, if we know that a number is divisible by both a and b, then it must be divisible by the least common multiple of a and b. For a = 6 and b = 2, the least common multiple is 6.

<< back to Problem B8


 

Problem B9

Refer to the divisibility tests for 2, 4, and 8. Since 10,000 is divisible by 16, we can test the last four digits of our given number. So, for example, to test whether the number 3,450,128 is divisible by 16, test the last four digits (0128) for divisibility by 16.

<< back to Problem B9


 

Problem B10

a. 

The test for divisibility by 12 encompasses the tests for 3 and 4. The sum of the digits must be a multiple of 3, and the last two digits must be a multiple of 4.

b. 

The test for divisibility by 18 encompasses the tests for 2 and 9. The units digit must be a multiple of 2, and the sum of the digits must be a multiple of 9.

c. 

The test for divisibility by 72 encompasses the tests for 8 and 9. The last three digits must be a multiple of 8, and the sum of the digits must be a multiple of 9.

<< back to Problem B10


 

Problem B11

a. 

The test for divisibility by 3 in base four is equivalent to the test for 9 in base ten: Add the digits of the number and check whether the sum is a multiple of 3. For example, the base four number 2313four is a multiple of 3, since the sum is 2 + 3 + 1 + 3 = 21four, which is a multiple of 3, since 3four • 3four = 21four.

You could also check this by converting 21four to base ten. In base ten, 21four equals 9, which is also a multiple of 3:

21four = (2 • 41) + (1 • 40) = 9

b. 

The test for divisibility by 2 in base five is equivalent to the test for 3 in base ten. The test is to add the digits of the number and then check whether the sum is a multiple of 2. For example, to check whether 42 five is divisible by 2, we add the digits, 4 + 2 = 11five. This number is divisible by 2, since 11five = 2five • 3five.

<< back to Problem B11


 

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