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Asif notices that $24^2 = 576$ and $2 + 4 = 6$ gives the last digit of 576 - AQA - A-Level Maths Pure - Question 6 - 2022 - Paper 2

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Asif notices that $24^2 = 576$ and $2 + 4 = 6$ gives the last digit of 576. He checks two more examples: $27^2 = 729$ and $2 + 7 = 9$ $29^2 = 841$ and $2 + 9 = 11... show full transcript

Worked Solution & Example Answer:Asif notices that $24^2 = 576$ and $2 + 4 = 6$ gives the last digit of 576 - AQA - A-Level Maths Pure - Question 6 - 2022 - Paper 2

Step 1

Give a counter example to show that Asif's conclusion is not correct.

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Answer

To counter Asif's conclusion, we will consider the square of 13. Calculating:

132=16913^2 = 169

Now, adding the digits of 13, we have:

1+3=41 + 3 = 4

The last digit of 169 is 9, while the sum of the digits of 13 gives us 4, which shows that Asif's method does not consistently give the correct last digit.

Step 2

Using Claire's method determine the last digit of 23456789^2.

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Answer

Using Claire's method, we only need the last digit of 23456789, which is 9. Calculating:

92=819^2 = 81

Therefore, the last digit of 23456789223456789^2 is 1.

Step 3

Given Claire's method is correct, use proof by exhaustion to show that no square number has a last digit of 8.

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Answer

To prove that no square number has a last digit of 8, we will consider the possible last digits of squares of integers.

The squares of integers can end in 0, 1, 4, 5, 6, or 9. Here are the calculations:

  • 02=00^2 = 0 (last digit 0)
  • 12=11^2 = 1 (last digit 1)
  • 22=42^2 = 4 (last digit 4)
  • 32=93^2 = 9 (last digit 9)
  • 42=64^2 = 6 (last digit 6)
  • 52=55^2 = 5 (last digit 5)
  • 62=66^2 = 6 (last digit 6)
  • 72=97^2 = 9 (last digit 9)
  • 82=48^2 = 4 (last digit 4)
  • 92=19^2 = 1 (last digit 1)

From the above, we can see that there is no case where the last digit is 8. Therefore, no square number can have a last digit of 8.

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