Investigations (Cambridge (CIE) IGCSE International Maths: Core): Exam Questions

Exam code: 0607

2 hours22 questions
1
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4 marks

This investigation looks at numbers that are divisible by 11, in particular exploring the patterns in the digits of these numbers.

Two-digit numbers that are divisible by 11 are easy to spot:

11, 22, 33, 44, 55, 66, 77, 88 and 99

However, three (or more) digit numbers that are divisible by 11, such as 319 and 28347, are harder to spot.

One method for testing if a number is divisible by 11 involves first calculating its alternating-digit sum, as follows:

(1st digit) - (2nd digit) + (3rd digit) - (4th digit) + ...

  • The 1st digit is the one farthest to the left

  • Digits are read from left to right

  • Alternating means the signs change from - to +

  • The first operation is always a subtraction, -

If the alternating-digit sum is a multiple of 11, then the original number is divisible by 11. The multiple of 11 can be:

  • positive, e.g. 11, 22, ...

  • negative, e.g. -11, -22, ...

  • or zero, 0

For example, to test if 319 is divisible by 11:

3 - 1 + 9 = 11

This is a multiple of 11, so 319 is divisible by 11.

Complete the table below. Parts of the table have been completed for you.

Number

Alternating digits

Sum

Multiple of 11?

573

5-7+3

1

No

253

2-5+3

0

Yes

18095

1-8+0-9+5

-11

Yes

698

5

3213

-1

8184

91949

50021

1-9+0-9+1-6

2
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4 marks

The following method is used to prove that a number, N, is a prime number:

  • Find square root of N

  • List all the prime numbers less than or equal to square root of N

  • Check if N is divisible by each of these prime numbers

The prime numbers 2, 3 and 5 have well-known divisibility tests that are often quicker to do than using a calculator (shown below). The divisibility test for 11 (from question 1) can now be added to this list.

This leaves divisibility by 7, 13, 17 and 19 to be done using a calculator.

For example, to prove that 313 is prime, check if it is divisible by all prime numbers less than or equal to square root of 313 equals 17.6918...

These are 2, 3, 5, 7, 11, 13 and 17:

Divisible by

Method

Working for 313

Conclusion

2

Last digit is even

3

No

3

Sum of digits is a multiple of 3

3+1+3 = 7

No

5

Last digit is 0 or 5

3

No

7

Calculator

313 over 7 equals 44.71...

No

11

Alternating-digit sum is a multiple of 11

3-1+3 = 5

No

13

Calculator

313 over 13 equals 24.07...

No

17

Calculator

313 over 17 equals 18.41...

No

19

Calculator

As 313 is not divisible by any prime number less than or equal to square root of 313, the conclusion is that 313 is a prime number.

The row for divisibility by 19 was not needed (since 19 greater than square root of 313).

Use the same table above to prove that the number 191 is a prime number.

Leave blank any rows that are not needed.

Divisible by

Method

Working for 191

Conclusion

2

Last digit is even

3

Sum of digits is a multiple of 3

5

Last digit is 0 or 5

7

Calculator

11

Alternating-digit sum is a multiple of 11

13

Calculator

17

Calculator

19

Calculator

3a
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1 mark

Digits can only take the integer values 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. A digit cannot be negative.

The five-digit number "6371 x" is divisible by 11.

Use the test in question 1 to find the value of the digit x.

3b
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2 marks

The five-digit number"19 a b 1" is divisible by 11 and has an alternating-digit sum of zero.

Use the table below to list all the possible values that a and b can take.

The first row has been done for you.

a

b

9

2

3c
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2 marks

If the five-digit number "19 a b 1" described in part (b) is also divisible by 3, find the only possible values of a and b.

Explain your answer.

4a
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1 mark

A four-digit number in the form "a a a a" is always divisible by 11. This is because the divisibility test in question 1 gives

a minus a plus a minus a equals 0

Show that a four-digit number in the form "a a b b" is always divisible by 11.

4b
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2 marks

A student believes that four-digit numbers in the form "a b a b" where a not equal to b are always divisible by 11.

Use an example of a number in the form "a b a b" and the divisibility test in question 1 to prove that the claim is not true.

4c
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4 marks

Determine which out of the following six-digit numbers are always divisible by 11.

You may assume that a not equal to b not equal to c.

The first row has been done for you.

Number

Working

Always divisible by 11?

a a a a a a

a minus a plus a minus a plus a minus a equals 0

Yes

a a b b c c

a b a b a b

a b b a a a

a b c a b c

4d
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2 marks

The number "a a a a b b b b" is an example of an eight-digit number that is always divisible by 11.

Using four digits of a and four digits of b, find two more examples of eight-digit numbers that are always divisible by 11.

Label the first digit as a.

5a
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1 mark

A number is called palindromic if it is equal to itself when the digits are written in reverse order.

For example, the numbers 33, 181, 9999, 12321 and 579975 are all palindromic.

Explain why the number 812128 is not palindromic.

5b
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2 marks

All palindromic numbers with an even number of digits are divisible by 11.

The table below can be used to prove this result for four, six and eight-digit palindromic numbers.

Complete the table.

The first row has been done for you.

Palindromic number

Working

Always divisible by 11?

a b b a

a minus b plus b minus a equals 0

Yes

a b c c b a

a b c d d c b a

5c
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2 marks

Palindromic numbers with an odd number of digits are not always divisible by 11.

Prove this statement using the five digit palindromic numbers 14641 and 13831.

6a
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1 mark

It is possible to form palindromic numbers by squaring multiples of 11.

Determine which, out of 11 squared, 22 squared and 33 squared, give palindromic numbers.

6b
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1 mark

Squaring larger multiples of 11 can produce larger palindromic numbers.

Find a six-digit palindromic number that is the square of a multiple of 11.

You may use the table below to help.

Multiple of 11

Multiple squared

11 cross times 73 equals 803

644809

11 cross times 74 equals 814

662596

11 cross times 75 equals 825

11 cross times 76 equals 836

11 cross times 77 equals 847

11 cross times 78 equals 858

736164

6c
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2 marks

It is possible to form palindromic numbers by adding together the squares of two different multiples of 11.

For example:

22 squared plus 88 squared equals 484 plus 7744 equals 8228

Find two more examples of palindromic numbers that are formed by adding together the squares of two different multiples of 11.

You may use the grid below to help.

Add (+)

22 squared

33 squared

33 squared

1573

44 squared

55 squared

66 squared

6d
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2 marks

A student says that the square of any positive multiple of 13 is never a palindromic number.

Investigate this claim and decide whether, or not, the student is correct.

Show your working clearly.

7a
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1 mark

A palindromic prime is a prime number that is also palindromic.

A two-digit palindromic prime is 11.

Use the information in question 2 to write down two different three-digit palindromic primes.

7b
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1 mark

Show that the number 111 is not a palindromic prime.

7c
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2 marks

Explain clearly why palindromic primes, other than 11, cannot have an even number of digits.

Use the information in question 5(b) to help.

8
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3 marks

A second test to determine whether or not a number is divisible by 11 is as follows:

  • Sum all the digits in the odd positions (1st digit + 3rd digit + 5th digit + ...)

  • Sum all the digits in the even positions (2nd digit + 4th digit + 6th digit +...)

  • Subtract the second sum from the first sum

  • If the result is a multiple of 11, the original number is divisible by 11

An eight-digit number is given by "a b c d e f g h".

Use algebra to prove that, for eight-digit numbers, this test is mathematically the same as the test in question 1.