DP Mathematics HL Questionbank

• Syllabus

10.6

Sub sections and their related questions

Fermat’s little theorem.

• 12M.3dm.hl.TZ0.5a: Use the result \(2003 = 6 \times 333 + 5\) and Fermat’s little theorem to show that...
• 12M.3dm.hl.TZ0.5b: Find \({2^{2003}}(\bmod 11)\) and \({2^{2003}}(\bmod 13)\).
• 12N.3dm.hl.TZ0.4a: State Fermat’s little theorem.
• 08M.3dm.hl.TZ2.3b: Show that \({x^{97}} - x + 1 \equiv 0(\bmod 97)\) has no solution.
• 08N.3dm.hl.TZ0.3c: Prove that \(\left. {22} \right|{5^{11}} + {17^{11}}\).
• 09M.3dm.hl.TZ0.5: (a)     Using Fermat’s little theorem, show that, in base 10, the last digit of n is always equal...
• 09N.3dm.hl.TZ0.4b: The representation of the positive integer N in base p is denoted by \({(N)_p}\) . If...
• SPNone.3dm.hl.TZ0.3a: One version of Fermat’s little theorem states that, under certain conditions,...
• 10M.3dm.hl.TZ0.1a: (i)     One version of Fermat’s little theorem states that, under certain...
• 10N.3dm.hl.TZ0.4: (a)     Write down Fermat’s little theorem. (b)     In base 5 the representation of a natural...
• 13M.3dm.hl.TZ0.5a: Show that \(\sum\limits_{k = 1}^p {{k^p} \equiv 0(\bmod p)} \).
• 13M.3dm.hl.TZ0.5b: Given that \(\sum\limits_{k = 1}^p {{k^{p - 1}} \equiv n(\bmod p)} \) where...
• 11N.3dm.hl.TZ0.5a: Use the above result to show that if p is prime then \({a^p} \equiv a(\bmod p)\) where a is any...
• 11N.3dm.hl.TZ0.5b: Show that \({2^{341}} \equiv 2(\bmod 341)\).
• 11N.3dm.hl.TZ0.5c: (i)     State the converse of the result in part (a). (ii)     Show that this converse is not true.
• 14N.3dm.hl.TZ0.1c: Explain why \(f(n)\) is always exactly divisible by \(5\).
• 14N.3dm.hl.TZ0.4b: Find the remainder when \({41^{82}}\) is divided by 11.
• 15N.3dm.hl.TZ0.3b: Hence, or otherwise, find the remainder when \({1982^{1982}}\) is divided by \(37\).
• 17N.3dm.hl.TZ0.2b: For every prime number \(p > 3\), show that \(p|{u_{p - 1}}\).
• 18M.3dm.hl.TZ0.2a: State Fermat’s little theorem.
• 18M.3dm.hl.TZ0.2b.i: Use Fermat’s little theorem to show that \(x \equiv {a^{p - 2}}b\left( {{\text{mod}}\,p} \right)\).