Hard GCSE Identities and argument Questions

Challenging, exam-style GCSE Identities and argument questions with worked solutions. Stretch yourself on the hardest proof, consecutive integers, multiples, odd numbers problems.

proofconsecutive integersmultiplesodd numbersidentitiescomparing coefficients
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
A student wants to prove that the product of two consecutive integers is always even. Which argument is fully correct?
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Worked solution

  1. Write two consecutive integers

    n,  n+1n, \; n+1

    Let n be any integer.

  2. Write their product

    n(n+1)n(n+1)

    Multiply the two integers.

  3. Consider whether n is even or odd

    n(n+1)n(n+1)

    Split into two cases.

  4. Case 1: n is even

    n=2kn = 2k

    Write n as 2 times an integer k.

  5. Then the product has a factor of 2

    2k(n+1)2k(n+1)

    The product is a multiple of 2.

  6. So the product is even

    2k(n+1)2k(n+1)

    An even factor makes the product even.

  7. Case 2: n is odd

    n=2k+1n = 2k + 1

    Write n as an odd number.

  8. Then n+1 is even

    n+1=2k+2=2(k+1)n + 1 = 2k + 2 = 2(k+1)

    One more than an odd number is even.

  9. So the product has a factor of 2

    n×2(k+1)n \times 2(k+1)

    Again the product is a multiple of 2.

  10. So the product is even

    n×2(k+1)n \times 2(k+1)

    An even factor makes the product even.

  11. Combine the cases

    n(n+1)n(n+1)

    In both cases the product is even.

  12. State the shortcut

    n or n+1 is evenn \text{ or } n+1 \text{ is even}

    Of any two consecutive integers, one is always even.

  13. Check n=4n = 4

    4×5=204 \times 5 = 20

    20 is even.

  14. Check n=7n = 7

    7×8=567 \times 8 = 56

    56 is even.

  15. Conclude

    n(n+1)n(n+1)

    So the product of two consecutive integers is always even.

Answer
n(n+1) is always even because one of two consecutive integers is even
Question 2
5 markschallenging
A two-digit number has tens digit aa and units digit bb, with a>ba > b. Prove that the difference between this number and the number formed by reversing its digits is always a multiple of 99.
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Worked solution

  1. Write the original number

    10a+b10a + b

    The value is 10 times the tens digit plus the units digit.

  2. Write the reversed number

    10b+a10b + a

    The digits swap places.

  3. Set up the difference

    (10a+b)−(10b+a)(10a + b) - (10b + a)

    Subtract the reversed number.

  4. Distribute the minus sign

    10a+b−10b−a10a + b - 10b - a

    Change the signs in the second bracket.

  5. Group the a terms

    10a−a10a - a

    Collect the terms in a.

  6. Add them

    9a9a

    10a minus a is 9a.

  7. Group the b terms

    b−10bb - 10b

    Collect the terms in b.

  8. Add them

    −9b-9b

    b minus 10b is -9b.

  9. Write the total

    9a−9b9a - 9b

    So the difference is 9a - 9b.

  10. Factor out 9

    9(a−b)9(a - b)

    Both terms share a factor of 9.

  11. Interpret

    9(a−b)9(a-b)

    9 times an integer is a multiple of 9.

  12. Note a - b is a positive integer

    a−ba - b

    Since a>ba > b, this is a whole number greater than 0.

  13. Check with 62 and 26

    62−26=36=9×462 - 26 = 36 = 9 \times 4

    36 is a multiple of 9.

  14. Check with 91 and 19

    91−19=72=9×891 - 19 = 72 = 9 \times 8

    72 is a multiple of 9.

  15. Conclude

    9(a−b)9(a-b)

    So the difference is always a multiple of 9.

Answer
9(a-b), a multiple of 9
Question 3
5 markschallenging
Which of the following is a complete and correct proof that the sum of three consecutive integers is always a multiple of 33?
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Worked solution

  1. Let the first integer be n

    n

    Take n as the smallest of the three.

  2. Write the second integer

    n+1n + 1

    One more than n.

  3. Write the third integer

    n+2n + 2

    Two more than n.

  4. Write the sum

    n+(n+1)+(n+2)n + (n+1) + (n+2)

    Add the three integers.

  5. Remove the brackets

    n+n+1+n+2n + n + 1 + n + 2

    Drop the brackets.

  6. Group the n terms

    n+n+nn + n + n

    Collect the terms in n.

  7. Add them

    3n3n

    There are three n's.

  8. Group the constants

    1+21 + 2

    Collect the number terms.

  9. Add them

    33

    1+2=31 + 2 = 3.

  10. Write the simplified sum

    3n+33n + 3

    So the sum is 3n + 3.

  11. Factor out 3

    3(n+1)3(n + 1)

    Both terms share a factor of 3.

  12. Interpret

    3(n+1)3(n+1)

    3 times an integer is a multiple of 3.

  13. Check n=4n = 4

    4+5+6=15=3×54 + 5 + 6 = 15 = 3 \times 5

    15 is a multiple of 3.

  14. Note n was arbitrary

    3(n+1)3(n+1)

    The argument works for any starting integer.

  15. Conclude

    3(n+1)3(n+1)

    So the sum is always a multiple of 3.

Answer
3(n+1), a multiple of 3
Question 4
6 markschallenging
A two-digit number has tens digit aa and units digit bb. Prove that the sum of this number and the number formed by reversing its digits is always a multiple of 1111.
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Worked solution

  1. Write the original number in terms of its digits

    10a+b10a + b

    The value is 10 times the tens digit plus the units digit.

  2. Explain the place value

    10a+b10a + b

    a tens and b units.

  3. Write the reversed number

    10b+a10b + a

    The digits swap places.

  4. Explain the reversal

    10b+a10b + a

    b is now the tens digit and a the units digit.

  5. Set up the sum

    (10a+b)+(10b+a)(10a + b) + (10b + a)

    Add the two numbers.

  6. Remove the brackets

    10a+b+10b+a10a + b + 10b + a

    Drop the brackets.

  7. Group the a terms

    10a+a10a + a

    Collect the terms in a.

  8. Add them

    11a11a

    10a plus a is 11a.

  9. Group the b terms

    b+10bb + 10b

    Collect the terms in b.

  10. Add them

    11b11b

    b plus 10b is 11b.

  11. Write the total

    11a+11b11a + 11b

    So the sum is 11a + 11b.

  12. Factor out 11

    11(a+b)11(a + b)

    Both terms share a factor of 11.

  13. Interpret

    11(a+b)11(a+b)

    11 times an integer is a multiple of 11.

  14. Check with 34 and 43

    34+43=77=11×734 + 43 = 77 = 11 \times 7

    77 is a multiple of 11.

  15. Conclude

    11(a+b)11(a+b)

    So the sum is always a multiple of 11.

Answer
11(a+b), a multiple of 11
Question 5
5 markschallenging
Show that (x+2)(x+3)−(x+1)(x+4)≡2(x+2)(x+3) - (x+1)(x+4) \equiv 2.
Show worked solution

Worked solution

  1. Expand the first product

    (x+2)(x+3)(x+2)(x+3)

    Multiply out the first pair of brackets.

  2. Collect the first product

    x2+5x+6x^2 + 5x + 6

    So (x+2)(x+3)=x2+5x+6(x+2)(x+3) = x^2 + 5x + 6.

  3. Expand the second product

    (x+1)(x+4)(x+1)(x+4)

    Multiply out the second pair of brackets.

  4. Collect the second product

    x2+5x+4x^2 + 5x + 4

    So (x+1)(x+4)=x2+5x+4(x+1)(x+4) = x^2 + 5x + 4.

  5. Set up the subtraction

    (x2+5x+6)−(x2+5x+4)(x^2 + 5x + 6) - (x^2 + 5x + 4)

    Subtract the second from the first.

  6. Distribute the minus sign

    x2+5x+6−x2−5x−4x^2 + 5x + 6 - x^2 - 5x - 4

    Change the signs in the second bracket.

  7. Cancel the x2x^2 terms

    5x+6−5x−45x + 6 - 5x - 4

    x2x^2 and −x2-x^2 cancel.

  8. Cancel the x terms

    6−46 - 4

    5x and -5x cancel.

  9. Subtract the constants

    22

    6 minus 4 is 2.

  10. Note the variable has vanished

    22

    The result no longer contains x.

  11. Compare with the right side

    2=22 = 2

    This matches.

  12. Check x=0x = 0

    6−4=26 - 4 = 2

    The identity holds at x=0x = 0.

  13. Check x=1x = 1

    12−10=212 - 10 = 2

    The identity holds at x=1x = 1.

  14. Explain

    22

    The expression equals 2 for every value of x.

  15. Conclude

    22

    So the identity holds for all x.

Answer
22

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